{"id":544851,"date":"2021-10-02T10:21:41","date_gmt":"2021-10-02T10:21:41","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=544851"},"modified":"2021-10-04T06:43:00","modified_gmt":"2021-10-04T06:43:00","slug":"rd-sharma-solutions-for-class-10-maths-chapter-9-arithmetic-progressions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-9-arithmetic-progressions\/","title":{"rendered":"RD Sharma Solutions for Class 10 Maths Chapter 9\u2013Arithmetic Progressions"},"content":{"rendered":"\n

Class 10: Maths Chapter 1 solutions. Complete Class 10 Maths Chapter 1 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 10 Maths Chapter 9\u2013Arithmetic Progressions<\/h2>\n\n\n\n

RD Sharma 10th Maths Chapter 9, Class 10 Maths Chapter 1 solutions<\/p>\n\n\n\n

Exercise 9.1 Page No: 9.5<\/h3>\n\n\n\n

1. Write the first terms of each of the following sequences whose nth<\/sup> term are:<\/strong><\/p>\n\n\n\n

(i) an<\/sub> = 3n + 2<\/strong><\/p>\n\n\n\n

(ii) an<\/sub> = (n \u2013 2)\/3<\/strong><\/p>\n\n\n\n

(iii) an<\/sub> = 3n<\/sup><\/strong><\/p>\n\n\n\n

(iv) an<\/sub> = (3n \u2013 2)\/ 5<\/strong><\/p>\n\n\n\n

(v) an<\/sub> = (-1)n <\/sup>. 2n<\/sup><\/strong><\/p>\n\n\n\n

(vi) an<\/sub> = n(n \u2013 2)\/2<\/strong><\/p>\n\n\n\n

(vii) an<\/sub> = n2<\/sup> \u2013 n + 1<\/strong><\/p>\n\n\n\n

(viii) an<\/sub> = n2<\/sup> \u2013 n + 1<\/strong><\/p>\n\n\n\n

(ix) an<\/sub> = (2n \u2013 3)\/ 6<\/strong><\/p>\n\n\n\n

Solutions:<\/strong><\/p>\n\n\n\n

(i) an<\/sub> = 3n + 2<\/p>\n\n\n\n

Given sequence whose an<\/sub> = 3n + 2<\/p>\n\n\n\n

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 and we get<\/p>\n\n\n\n

a1<\/sub> = (3 \u00d7 1) + 2 = 3 + 2 = 5<\/p>\n\n\n\n

a2<\/sub> = (3 \u00d7 2) + 2 = 6 + 2 = 8<\/p>\n\n\n\n

a3<\/sub> = (3 \u00d7 3) + 2 = 9 + 2 = 11<\/p>\n\n\n\n

a4<\/sub> = (3 \u00d7 4) + 2 = 12 + 2 = 14<\/p>\n\n\n\n

a5<\/sub> = (3 \u00d7 5) + 2 = 15 + 2 = 17<\/p>\n\n\n\n

\u2234 the required first five terms of the sequence whose nth<\/sup> term,<\/strong> an<\/sub> = 3n + 2 are 5, 8, 11, 14, 17.<\/p>\n\n\n\n

(ii) an<\/sub> = (n \u2013 2)\/3<\/p>\n\n\n\n

Given sequence whose <\/strong><\/p>\n\n\n\n

On putting n = 1, 2, 3, 4, 5 then can get the first five terms<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

\u2234 the required first five terms of the sequence whose nth<\/sup> term,<\/strong><\/p>\n\n\n\n

(iii) an<\/sub> = 3n<\/sup><\/p>\n\n\n\n

Given sequence whose an<\/sub> = 3n<\/sup><\/p>\n\n\n\n

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 in the above<\/p>\n\n\n\n

a1<\/sub> = 31<\/sup> = 3;<\/p>\n\n\n\n

a2<\/sub> = 32<\/sup> = 9;<\/p>\n\n\n\n

a3<\/sub> = 27;<\/p>\n\n\n\n

a4<\/sub> = 34<\/sup> = 81;<\/p>\n\n\n\n

a5<\/sub> = 35<\/sup> = 243.<\/p>\n\n\n\n

\u2234 the required first five terms of the sequence whose nth<\/sup> term,<\/strong> an<\/sub> = 3n<\/sup> are 3, 9, 27, 81, 243.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv) an<\/sub> = (3n \u2013 2)\/ 5<\/p>\n\n\n\n

Given sequence whose<\/p>\n\n\n\n

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above<\/p>\n\n\n\n

And, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 the required first five terms of the sequence are 1\/5, 4\/5, 7\/5, 10\/5, 13\/5<\/p>\n\n\n\n

(v) an<\/sub> = (-1)n<\/sup>2n<\/sup><\/p>\n\n\n\n

Given sequence whose an<\/sub> = (-1)n<\/sup>2n<\/sup><\/p>\n\n\n\n

To get first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above.<\/p>\n\n\n\n

a1<\/sub> = (-1)1<\/sup>.21<\/sup> = (-1).2 = -2<\/p>\n\n\n\n

a2<\/sub> = (-1)2<\/sup>.22<\/sup> = (-1).4 = 4<\/p>\n\n\n\n

a3<\/sub> = (-1)3<\/sup>.23<\/sup> = (-1).8 = -8<\/p>\n\n\n\n

a4<\/sub> = (-1)4<\/sup>.24<\/sup> = (-1).16 = 16<\/p>\n\n\n\n

a5<\/sub> = (-1)5<\/sup>.25<\/sup> = (-1).32 = -32<\/p>\n\n\n\n

\u2234 the first five terms of the sequence are \u2013 2, 4, \u2013 8, 16, \u2013 32.<\/p>\n\n\n\n

(vi) an<\/sub> = n(n \u2013 2)\/2<\/p>\n\n\n\n

The given sequence is,
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.<\/p>\n\n\n\n

And, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 the required first five terms are -1\/2, 0, 3\/2, 4, 15\/2<\/p>\n\n\n\n

(vii) an<\/sub> = n2<\/sup> \u2013 n + 1<\/p>\n\n\n\n

The given sequence whose, an<\/sub> = n2<\/sup> \u2013 n + 1<\/p>\n\n\n\n

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5.<\/p>\n\n\n\n

And, we get<\/p>\n\n\n\n

a1<\/sub> = 12<\/sup> \u2013 1 + 1 = 1<\/p>\n\n\n\n

a2<\/sub> = 22<\/sup> \u2013 2 + 1 = 3<\/p>\n\n\n\n

a3<\/sub> = 32<\/sup> \u2013 3 + 1 = 7<\/p>\n\n\n\n

a4<\/sub> = 42<\/sup> \u2013 4 + 1 = 13<\/p>\n\n\n\n

a5<\/sub> = 52<\/sup> \u2013 5 + 1 = 21<\/p>\n\n\n\n

\u2234 the required first five terms of the sequence are 1, 3, 7, 13, 21.<\/p>\n\n\n\n

(viii) an<\/sub> = 2n2<\/sup> \u2013 3n + 1<\/p>\n\n\n\n

The given sequence whose an<\/sub> = 2n2<\/sup> \u2013 3n + 1<\/p>\n\n\n\n

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.<\/p>\n\n\n\n

And, we get<\/p>\n\n\n\n

a1<\/sub> = 2.12<\/sup> \u2013 3.1 + 1 = 2 \u2013 3 + 1 = 0<\/p>\n\n\n\n

a2<\/sub> = 2.22<\/sup> \u2013 3.2 + 1 = 8 \u2013 6 + 1 = 3<\/p>\n\n\n\n

a3<\/sub> = 2.32<\/sup> \u2013 3.3 + 1 = 18 \u2013 9 + 1 = 10<\/p>\n\n\n\n

a4<\/sub> = 2.42<\/sup> \u2013 3.4 + 1 = 32 \u2013 12 + 1 = 21<\/p>\n\n\n\n

a5<\/sub> = 2.52<\/sup> \u2013 3.5 + 1 = 50 \u2013 15 + 1 = 36<\/p>\n\n\n\n

\u2234 the required first five terms of the sequence are 0, 3, 10, 21, 36.<\/p>\n\n\n\n

(ix) an<\/sub> = (2n \u2013 3)\/ 6<\/p>\n\n\n\n

Given sequence whose,
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

To get the first five terms of the sequence we put n = 1, 2, 3, 4, 5.<\/p>\n\n\n\n

And, we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 the required first five terms of the sequence are -1\/6, 1\/6, 1\/2, 5\/6 and 7\/6<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.2 Page No: 9.8<\/h4>\n\n\n\n

1. Show that the sequence defined by an<\/sub> = 5n \u2013 7 is an A.P., find its common difference.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, an<\/sub> = 5n \u2013 7<\/p>\n\n\n\n

Now putting n = 1, 2, 3, 4 we get,<\/p>\n\n\n\n

a1<\/sub> = 5(1) \u2013 7 = 5 \u2013 7 = -2<\/p>\n\n\n\n

a2<\/sub> = 5(2) \u2013 7 = 10 \u2013 7 = 3<\/p>\n\n\n\n

a3<\/sub> = 5(3) \u2013 7 = 15 \u2013 7 = 8<\/p>\n\n\n\n

a4<\/sub> = 5(4) \u2013 7 = 20 \u2013 7 = 13<\/p>\n\n\n\n

We can see that,<\/p>\n\n\n\n

a2<\/sub> \u2013 a1 <\/sub>= 3 \u2013 (-2) = 5<\/p>\n\n\n\n

a3<\/sub> \u2013 a2 <\/sub>= 8 \u2013 (3) = 5<\/p>\n\n\n\n

a4<\/sub> \u2013 a3 <\/sub>= 13 \u2013 (8) = 5<\/p>\n\n\n\n

Since the difference between the terms is common, we can conclude that the given sequence defined by an<\/sub> = 5n \u2013 7 is an A.P with common difference 5.<\/p>\n\n\n\n

2. Show that the sequence defined by an<\/sub> = 3n2<\/sup> \u2013 5 is not an A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, an<\/sub> = 3n2<\/sup> \u2013 5<\/p>\n\n\n\n

Now putting n = 1, 2, 3, 4 we get,<\/p>\n\n\n\n

a1<\/sub> = 3(1)2<\/sup> \u2013 5= 3 \u2013 5 = -2<\/p>\n\n\n\n

a2<\/sub> = 3(2)2<\/sup> \u2013 5 = 12 \u2013 5 = 7<\/p>\n\n\n\n

a3<\/sub> = 3(3)2<\/sup> \u2013 5 = 27 \u2013 5 = 22<\/p>\n\n\n\n

a4<\/sub> = 3(4)2<\/sup> \u2013 5 = 48 \u2013 5 = 43<\/p>\n\n\n\n

We can see that,<\/p>\n\n\n\n

a2<\/sub> \u2013 a1 <\/sub>= 7 \u2013 (-2) = 9<\/p>\n\n\n\n

a3<\/sub> \u2013 a2 <\/sub>= 22 \u2013 7 = 15<\/p>\n\n\n\n

a4<\/sub> \u2013 a3 <\/sub>= 43 \u2013 22 = 21<\/p>\n\n\n\n

Since the difference between the terms is not common and varying, we can conclude that the given sequence defined by an<\/sub> = 3n2<\/sup> \u2013 5 is not an A.P.<\/p>\n\n\n\n

3. The general term of a sequence is given by an<\/sub> = -4n + 15. Is the sequence an A.P.? If so, find its 15th<\/sup> term and the common difference.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, an<\/sub> = -4n + 15<\/p>\n\n\n\n

Now putting n = 1, 2, 3, 4 we get,<\/p>\n\n\n\n

a1<\/sub> = -4(1) + 15 = -4 + 15 = 11<\/p>\n\n\n\n

a2<\/sub> = -4(2) + 15 = -8 + 15 = 7<\/p>\n\n\n\n

a3<\/sub> = -4(3) + 15 = -12 + 15 = 3<\/p>\n\n\n\n

a4<\/sub> = -4(4) + 15 = -16 + 15 = -1<\/p>\n\n\n\n

We can see that,<\/p>\n\n\n\n

a2<\/sub> \u2013 a1 <\/sub>= 7 \u2013 (11) = -4<\/p>\n\n\n\n

a3<\/sub> \u2013 a2 <\/sub>= 3 \u2013 7 = -4<\/p>\n\n\n\n

a4<\/sub> \u2013 a3 <\/sub>= -1 \u2013 3 = -4<\/p>\n\n\n\n

Since the difference between the terms is common, we can conclude that the given sequence defined by an<\/sub> = -4n + 15 is an A.P with common difference of -4.<\/p>\n\n\n\n

Hence, the 15th<\/sup> term will be<\/p>\n\n\n\n

a15<\/sub> = -4(15) + 15 = -60 + 15 = -45<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.3 Page No: 9.11<\/h4>\n\n\n\n

1. For the following arithmetic progressions write the first term a and the common difference d:<\/strong><\/p>\n\n\n\n

(i) \u2013 5, -1, 3, 7,\u2026<\/strong><\/p>\n\n\n\n

(ii) 1\/5, 3\/5, 5\/5, 7\/5,\u2026<\/strong><\/p>\n\n\n\n

(iii) 0.3, 0.55, 0.80, 1.05,\u2026<\/strong><\/p>\n\n\n\n

(iv) -1.1, \u2013 3.1, \u2013 5.1, \u2013 7.1,\u2026<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that if a is the first term and d is the common difference, the arithmetic progression is a, a + d, a + 2d + a + 3d,\u2026.<\/p>\n\n\n\n

(i) \u2013 5, \u20131, 3, 7,\u2026<\/p>\n\n\n\n

Given arithmetic series is \u2013 5, \u20131, 3, 7,\u2026<\/p>\n\n\n\n

c a, a + d, a + 2d + a + 3d,\u2026.<\/p>\n\n\n\n

Thus, by comparing these two we get, a = \u2013 5, a + d = 1, a + 2d = 3, a + 3d = 7<\/p>\n\n\n\n

First term (a) = \u2013 5<\/p>\n\n\n\n

By subtracting second and first term, we get<\/p>\n\n\n\n

(a + d) \u2013 (a) = d<\/p>\n\n\n\n

-1 \u2013 (- 5) = d<\/p>\n\n\n\n

4 = d<\/p>\n\n\n\n

\u21d2 Common difference (d) = 4.<\/p>\n\n\n\n

(ii) 1\/5, 3\/5, 5\/5, 7\/5, \u2026\u2026\u2026\u2026.<\/p>\n\n\n\n

Given arithmetic series is 1\/5, 3\/5, 5\/5, 7\/5, \u2026\u2026\u2026\u2026\u2026<\/p>\n\n\n\n

It is seen that, it\u2019s of the form of 1\/5, 2\/5, 5\/5, 7\/5, \u2026\u2026\u2026.. a, a + d, a + 2d, a + 3d,<\/p>\n\n\n\n

Thus, by comparing these two, we get<\/p>\n\n\n\n

a = 1\/5, a + d = 3\/5, a + 2d = 5\/5, a + 3d = 7\/5<\/p>\n\n\n\n

First term (a) = 1\/5<\/p>\n\n\n\n

By subtracting first term from second term, we get<\/p>\n\n\n\n

d = (a + d)-(a)<\/p>\n\n\n\n

d = 3\/5 \u2013 1\/5<\/p>\n\n\n\n

d = 2\/5<\/p>\n\n\n\n

\u21d2 common difference (d) = 2\/5<\/p>\n\n\n\n

(iii) 0.3, 0.55, 0.80, 1.05, \u2026\u2026\u2026\u2026<\/p>\n\n\n\n

Given arithmetic series 0.3, 0.55, 0.80, 1.05, \u2026\u2026\u2026.<\/p>\n\n\n\n

It is seen that, it\u2019s of the form of a, a + d, a + 2d, a + 3d,<\/p>\n\n\n\n

Thus, by comparing we get,<\/p>\n\n\n\n

a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05<\/p>\n\n\n\n

First term (a) = 0.3.<\/p>\n\n\n\n

By subtracting first term from second term. We get<\/p>\n\n\n\n

d = (a + d) \u2013 (a)<\/p>\n\n\n\n

d = 0.55 \u2013 0.3<\/p>\n\n\n\n

d = 0.25<\/p>\n\n\n\n

\u21d2 Common difference (d) = 0.25<\/p>\n\n\n\n

(iv) \u20131.1, \u2013 3.1, \u2013 5.1, \u20137.1, \u2026\u2026..<\/p>\n\n\n\n

General series is \u20131.1, \u2013 3.1, \u2013 5.1, \u20137.1, \u2026\u2026..<\/p>\n\n\n\n

It is seen that, it\u2019s of the form of a, a + d, a + 2d, a + 3d, \u2026\u2026\u2026..<\/p>\n\n\n\n

Thus, by comparing these two, we get<\/p>\n\n\n\n

a = \u20131.1, a + d = \u20133.1, a + 2d = \u20135.1, a + 3d = \u20137.1<\/p>\n\n\n\n

First term (a) = \u20131.1<\/p>\n\n\n\n

Common difference (d) = (a + d) \u2013 (a)<\/p>\n\n\n\n

= -3.1 \u2013 ( \u2013 1.1)<\/p>\n\n\n\n

\u21d2 Common difference (d) = \u2013 2<\/p>\n\n\n\n

2. Write the arithmetic progression when first term a and common difference d are as follows:<\/strong><\/p>\n\n\n\n

(i) a = 4, d = \u2013 3<\/strong><\/p>\n\n\n\n

(ii) a = \u20131, d = 1\/2<\/strong><\/p>\n\n\n\n

(iii) a = \u20131.5, d = \u2013 0.5<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that, if first term (a) = a and common difference = d, then the arithmetic series is: a, a + d, a + 2d, a + 3d,<\/p>\n\n\n\n

(i) a = 4, d = -3<\/p>\n\n\n\n

Given, first term (a) = 4<\/p>\n\n\n\n

Common difference (d) = -3<\/p>\n\n\n\n

Then arithmetic progression is: a, a + d, a + 2d, a + 3d, \u2026\u2026<\/p>\n\n\n\n

\u21d2 4, 4 \u2013 3, 4 + 2(-3), 4 + 3(-3), \u2026\u2026<\/p>\n\n\n\n

\u21d2 4, 1, \u2013 2, \u2013 5, \u2013 8 \u2026\u2026..<\/p>\n\n\n\n

(ii) a = -1, d = 1\/2<\/p>\n\n\n\n

Given, first term (a) = -1<\/p>\n\n\n\n

Common difference (d) = 1\/2<\/p>\n\n\n\n

Then arithmetic progression is: a, a + d, a + 2d, a + 3d,<\/p>\n\n\n\n

\u21d2 -1, -1 + 1\/2, -1 + 2\u00bd, -1 + 3\u00bd, \u2026<\/p>\n\n\n\n

\u21d2 -1, -1\/2, 0, 1\/2<\/p>\n\n\n\n

(iii) a = \u20131.5, d = \u2013 0.5<\/p>\n\n\n\n

Given First term (a) = \u20131.5<\/p>\n\n\n\n

Common difference (d) = \u2013 0.5<\/p>\n\n\n\n

Then arithmetic progression is; a, a + d, a + 2d, a + 3d, \u2026\u2026<\/p>\n\n\n\n

\u21d2 -1.5, -1.5 + (-0.5), \u20131.5 + 2(\u2013 0.5), \u20131.5 + 3(\u2013 0.5)<\/p>\n\n\n\n

\u21d2 \u2013 1.5, \u2013 2, \u2013 2.5, \u2013 3, \u2026\u2026.<\/p>\n\n\n\n

3. In which of the following situations, the sequence of numbers formed will form an A.P.?
(i) The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.
(ii) The amount of air present in the cylinder when a vacuum pump removes each time 1\/4 of their remaining in the cylinder. <\/strong><\/p>\n\n\n\n

(iii) Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, \u2026, and so on.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given,<\/p>\n\n\n\n

Cost of digging a well for the first meter (c1<\/sub>) = Rs.150.<\/p>\n\n\n\n

And, the cost rises by Rs.20 for each succeeding meter<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Cost of digging for the second meter (c2<\/sub>) = Rs.150 + Rs 20 = Rs 170<\/p>\n\n\n\n

Cost of digging for the third meter (c3<\/sub>) = Rs.170 + Rs 20 = Rs 210<\/p>\n\n\n\n

Hence, its clearly seen that the costs of digging a well for different lengths are 150, 170, 190, 210, \u2026.<\/p>\n\n\n\n

Evidently, this series is in A\u2219P.<\/p>\n\n\n\n

With first term (a) = 150, common difference (d) = 20<\/p>\n\n\n\n

(ii) Given,<\/p>\n\n\n\n

Let the initial volume of air in a cylinder be V liters each time 3th<\/sup>\/4 of air in a remaining i.e<\/p>\n\n\n\n

1 -1\/4<\/p>\n\n\n\n

First time, the air in cylinder is V.<\/p>\n\n\n\n

Second time, the air in cylinder is 3\/4 V.<\/p>\n\n\n\n

Third time, the air in cylinder is (3\/4)2<\/sup> V.<\/p>\n\n\n\n

Thus, series is V, 3\/4 V, (3\/4)2 <\/sup>V,(3\/4)3 <\/sup>V, \u2026.<\/p>\n\n\n\n

Hence, the above series is not a A.P.<\/p>\n\n\n\n

(iii) Given,<\/p>\n\n\n\n

Divya deposited Rs 1000 at compound interest of 10% p.a<\/p>\n\n\n\n

So, the amount at the end of first year is = 1000 + 0.1(1000) = Rs 1100<\/p>\n\n\n\n

And, the amount at the end of second year is = 1100 + 0.1(1100) = Rs 1210<\/p>\n\n\n\n

And, the amount at the end of third year is = 1210 + 0.1(1210) = Rs 1331<\/p>\n\n\n\n

Cleary, these amounts 1100, 1210 and 1331 are not in an A.P since the difference between them is not the same.<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.4 Page No: 9.24<\/h4>\n\n\n\n

1. Find:<\/strong><\/p>\n\n\n\n

(i) 10th<\/sup> tent of the AP 1, 4, 7, 10\u2026.<\/strong><\/p>\n\n\n\n

(ii) 18th<\/sup> term of the AP \u221a2, 3\u221a2, 5\u221a2, \u2026\u2026. <\/strong><\/p>\n\n\n\n

(iii) nth<\/sup> term of the AP 13, 8, 3, -2, \u2026\u2026\u2026. <\/strong><\/p>\n\n\n\n

(iv) 10th<\/sup> term of the AP -40, -15, 10, 35, \u2026\u2026\u2026\u2026. <\/strong><\/p>\n\n\n\n

(v) 8th<\/sup> term of the AP 11, 104, 91, 78, \u2026\u2026\u2026\u2026\u2026 <\/strong><\/p>\n\n\n\n

(vi) 11th<\/sup> tenor of the AP 10.0, 10.5, 11.0, 11.2, \u2026\u2026\u2026\u2026.. <\/strong><\/p>\n\n\n\n

(vii) 9th<\/sup> term of the AP 3\/4, 5\/4, 7\/4 + 9\/4, \u2026\u2026\u2026..<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i<\/strong>) Given A.P. is 1, 4, 7, 10, \u2026\u2026\u2026.<\/p>\n\n\n\n

First term (a) = 1<\/p>\n\n\n\n

Common difference (d) = Second term \u2013 First term<\/p>\n\n\n\n

= 4 \u2013 1 = 3.<\/p>\n\n\n\n

We know that, nth<\/sup> term in an A.P = a + (n \u2013 1)d<\/p>\n\n\n\n

Then, 10th<\/sup> term in the A.P is 1 + (10 \u2013 1)3<\/p>\n\n\n\n

= 1 + 9\u00d73<\/p>\n\n\n\n

= 1 + 27<\/p>\n\n\n\n

= 28<\/p>\n\n\n\n

\u2234 10th<\/sup> term of A. P. is 28<\/p>\n\n\n\n

(ii) Given A.P. is \u221a2, 3\u221a2, 5\u221a2, \u2026\u2026.<\/p>\n\n\n\n

First term (a) = \u221a2<\/p>\n\n\n\n

Common difference = Second term \u2013 First term<\/p>\n\n\n\n

= 3\u221a2 \u2013 \u221a2<\/p>\n\n\n\n

\u21d2 d = 2\u221a2<\/p>\n\n\n\n

We know that, nth<\/sup> term in an A. P. = a + (n \u2013 1)d<\/p>\n\n\n\n

Then, 18th<\/sup> term of A. P. = \u221a2 + (18 \u2013 1)2\u221a2<\/p>\n\n\n\n

= \u221a2 + 17.2\u221a2<\/p>\n\n\n\n

= \u221a2 (1+34)<\/p>\n\n\n\n

= 35\u221a2<\/p>\n\n\n\n

\u2234 18th<\/sup> term of A. P. is 35\u221a2<\/p>\n\n\n\n

(iii) Given A. P. is 13, 8, 3, \u2013 2,  \u2026\u2026\u2026\u2026<\/p>\n\n\n\n

First term (a) = 13<\/p>\n\n\n\n

Common difference (d) = Second term first term<\/p>\n\n\n\n

= 8 \u2013 13 = \u2013 5<\/p>\n\n\n\n

We know that, nth<\/sup> term of an A.P. an<\/sub> = a +(n \u2013 1)d<\/p>\n\n\n\n

= 13 + (n \u2013 1) \u2013 5<\/p>\n\n\n\n

= 13 \u2013 5n + 5<\/p>\n\n\n\n

\u2234 nth<\/sup> term of the A.P is an<\/sub> = 18 \u2013 5n<\/p>\n\n\n\n

(iv) Given A. P. is \u2013 40, -15, 10, 35, \u2026\u2026\u2026.<\/p>\n\n\n\n

First term (a) = -40<\/p>\n\n\n\n

Common difference (d) = Second term \u2013 fast term<\/p>\n\n\n\n

= -15 \u2013 (- 40)<\/p>\n\n\n\n

= 40 \u2013 15<\/p>\n\n\n\n

= 25<\/p>\n\n\n\n

We know that, nth<\/sup> term of an A.P. an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Then, 10th<\/sup> term of A. P. a10<\/sub> = -40 + (10 \u2013 1)25<\/p>\n\n\n\n

= \u2013 40 + 9.25<\/p>\n\n\n\n

= \u2013 40 + 225<\/p>\n\n\n\n

= 185<\/p>\n\n\n\n

\u2234 10th<\/sup> term of the A. P. is 185<\/p>\n\n\n\n

(v) Given sequence is 117, 104, 91, 78, \u2026\u2026\u2026\u2026.<\/p>\n\n\n\n

First term (a) = 117<\/p>\n\n\n\n

Common difference (d) = Second term \u2013 first term<\/p>\n\n\n\n

= 104 \u2013 117<\/p>\n\n\n\n

= \u2013 13<\/p>\n\n\n\n

We know that, nth<\/sup> term = a + (n \u2013 1)d<\/p>\n\n\n\n

Then, 8th<\/sup> term = a + (8 \u2013 1)d<\/p>\n\n\n\n

= 117 + 7(-13)<\/p>\n\n\n\n

= 117 \u2013 91<\/p>\n\n\n\n

= 26<\/p>\n\n\n\n

\u2234 8th<\/sup> term of the A. P. is 26<\/p>\n\n\n\n

(vi) Given A. P is 10.0, 10.5, 11.0, 11.5,<\/p>\n\n\n\n

First term (a) = 10.0<\/p>\n\n\n\n

Common difference (d) = Second term \u2013 first term<\/p>\n\n\n\n

= 10.5 \u2013 10.0 = 0.5<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Then, 11th<\/sup> term a11<\/sub> = 10.0 + (11 \u2013 1)0.5<\/p>\n\n\n\n

= 10.0 + 10 x 0.5<\/p>\n\n\n\n

= 10.0 + 5<\/p>\n\n\n\n

=15.0<\/p>\n\n\n\n

\u2234 11th<\/sup> term of the A. P. is 15.0<\/p>\n\n\n\n

(vii) Given A. P is 3\/4,  5\/4, 7\/4, 9\/4, \u2026\u2026\u2026\u2026<\/p>\n\n\n\n

First term (a) = 3\/4<\/p>\n\n\n\n

Common difference (d) = Second term \u2013 first term<\/p>\n\n\n\n

= 5\/4 \u2013 3\/4<\/p>\n\n\n\n

= 2\/4<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Then, 9th<\/sup> term a9<\/sub> = a + (9 \u2013 1)d<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2234 9th<\/sup> term of the A. P. is 19\/4.<\/p>\n\n\n\n

2.(i) Which term of the AP 3, 8, 13, \u2026. is 248?<\/strong><\/p>\n\n\n\n

(ii) Which term of the AP 84, 80, 76, \u2026 is 0?<\/strong><\/p>\n\n\n\n

(iii) Which term of the AP 4. 9, 14, \u2026. is 254?<\/strong><\/p>\n\n\n\n

(iv) Which term of the AP 21. 42, 63, 84, \u2026 is 420?<\/strong><\/p>\n\n\n\n

(v) Which term of the AP 121, 117. 113, \u2026 is its first negative term?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given A.P. is 3, 8, 13, \u2026\u2026\u2026..<\/p>\n\n\n\n

First term (a) = 3<\/p>\n\n\n\n

Common difference (d) = Second term \u2013 first term<\/p>\n\n\n\n

= 8 \u2013 3<\/p>\n\n\n\n

= 5<\/p>\n\n\n\n

We know that, nth<\/sup> term (an<\/sub>) = a + (n \u2013 1)d<\/p>\n\n\n\n

And, given nth<\/sup> term an<\/sub> = 248<\/p>\n\n\n\n

248 = 3+(n \u2013 1)5<\/p>\n\n\n\n

248 = -2 + 5n<\/p>\n\n\n\n

5n = 250<\/p>\n\n\n\n

n =250\/5 = 50<\/p>\n\n\n\n

\u2234 50th<\/sup> term in the A.P is 248.<\/p>\n\n\n\n

(ii) Given A. P is 84, 80, 76, \u2026\u2026\u2026\u2026<\/p>\n\n\n\n

First term (a) = 84<\/p>\n\n\n\n

Common difference (d) = a2<\/sub> \u2013 a<\/p>\n\n\n\n

= 80 \u2013 84<\/p>\n\n\n\n

= \u2013 4<\/p>\n\n\n\n

We know that, nth<\/sup> term (an<\/sub>) = a +(n \u2013 1)d<\/p>\n\n\n\n

And, given nth term is 0<\/p>\n\n\n\n

0 = 84 + (n \u2013 1) \u2013 4<\/p>\n\n\n\n

84 = +4(n \u2013  1)<\/p>\n\n\n\n

n \u2013 1 = 84\/4 = 21<\/p>\n\n\n\n

n = 21 + 1 = 22<\/p>\n\n\n\n

\u2234 22nd<\/sup> term in the A.P is 0.<\/p>\n\n\n\n

(iii) Given A. P 4, 9, 14, \u2026\u2026\u2026\u2026<\/p>\n\n\n\n

First term (a) = 4<\/p>\n\n\n\n

Common difference (d) = a2<\/sub> \u2013 a1<\/sub><\/p>\n\n\n\n

= 9 \u2013 4<\/p>\n\n\n\n

= 5<\/p>\n\n\n\n

We know that, nth<\/sup> term (an<\/sub>) = a + (n \u2013 1)d<\/p>\n\n\n\n

And, given nth<\/sup> term is 254<\/p>\n\n\n\n

4 + (n \u2013 1)5 = 254<\/p>\n\n\n\n

(n \u2013 1)\u22195 = 250<\/p>\n\n\n\n

n \u2013 1 = 250\/5 = 50<\/p>\n\n\n\n

n = 51<\/p>\n\n\n\n

\u2234  51th<\/sup> term in the A.P is 254.<\/p>\n\n\n\n

(iv) Given A. P 21, 42, 63, 84, \u2026\u2026\u2026<\/p>\n\n\n\n

a = 21, d = a2<\/sub> \u2013 a1<\/sub><\/p>\n\n\n\n

= 42 \u2013 21<\/p>\n\n\n\n

= 21<\/p>\n\n\n\n

We know that, nth<\/sup> term (an<\/sub>) = a +(n \u2013 1)d<\/p>\n\n\n\n

And, given nth<\/sup> term = 420<\/p>\n\n\n\n

21 + (n \u2013 1)21 = 420<\/p>\n\n\n\n

(n \u2013 1)21 = 399<\/p>\n\n\n\n

n \u2013 1 = 399\/21 = 19<\/p>\n\n\n\n

n = 20<\/p>\n\n\n\n

\u2234 20th<\/sup> term is 420.<\/p>\n\n\n\n

(v) Given A.P is 121, 117, 113, \u2026\u2026\u2026..<\/p>\n\n\n\n

Fiat term (a) = 121<\/p>\n\n\n\n

Common difference (d) = 117 \u2013 121<\/p>\n\n\n\n

= \u2013 4<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

And, for some nth<\/sup> term is negative i.e., an<\/sub> < 0<\/p>\n\n\n\n

121 + (n \u2013 1) \u2013 4 < 0<\/p>\n\n\n\n

121 + 4 \u2013 4n < 0<\/p>\n\n\n\n

125 \u2013 4n < 0<\/p>\n\n\n\n

4n > 125<\/p>\n\n\n\n

n > 125\/4<\/p>\n\n\n\n

n > 31.25<\/p>\n\n\n\n

The integer which comes after 31.25 is 32.<\/p>\n\n\n\n

\u2234 32nd<\/sup> term in the A.P will be the first negative term.<\/p>\n\n\n\n

3.(i) Is 68 a term of the A.P. 7, 10, 13,\u2026 ?<\/strong><\/p>\n\n\n\n

(ii) Is 302 a term of the A.P. 3, 8, 13, \u2026. ?<\/strong><\/p>\n\n\n\n

(iii) Is -150 a term of the A.P. 11, 8, 5, 2, \u2026 ?<\/strong><\/p>\n\n\n\n

Solutions:<\/strong><\/p>\n\n\n\n

(i) Given, A.P. 7, 10, 13,\u2026<\/p>\n\n\n\n

Here, a = 7 and d = a2<\/sub> \u2013 a1<\/sub> = 10 \u2013 7 = 3<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Required to check nth<\/sup> term an<\/sub> = 68<\/p>\n\n\n\n

a + (n \u2013 1)d = 68<\/p>\n\n\n\n

7 + (n \u2013 1)3 = 68<\/p>\n\n\n\n

7 + 3n \u2013 3 = 68<\/p>\n\n\n\n

3n + 4 = 68<\/p>\n\n\n\n

3n = 64<\/p>\n\n\n\n

\u21d2 n = 64\/3, which is not a whole number.<\/p>\n\n\n\n

Therefore, 68 is not a term in the A.P.<\/p>\n\n\n\n

(ii) Given, A.P. 3, 8, 13,\u2026<\/p>\n\n\n\n

Here, a = 3 and d = a2<\/sub> \u2013 a1<\/sub> = 8 \u2013 3 = 5<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Required to check nth<\/sup> term an<\/sub> = 302<\/p>\n\n\n\n

a + (n \u2013 1)d = 302<\/p>\n\n\n\n

3 + (n \u2013 1)5 = 302<\/p>\n\n\n\n

3 + 5n \u2013 5 = 302<\/p>\n\n\n\n

5n \u2013 2 = 302<\/p>\n\n\n\n

5n = 304<\/p>\n\n\n\n

\u21d2 n = 304\/5, which is not a whole number.<\/p>\n\n\n\n

Therefore, 302 is not a term in the A.P.<\/p>\n\n\n\n

(iii) Given, A.P. 11, 8, 5, 2, \u2026<\/p>\n\n\n\n

Here, a = 11 and d = a2<\/sub> \u2013 a1<\/sub> = 8 \u2013 11 = -3<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Required to check nth<\/sup> term an<\/sub> = -150<\/p>\n\n\n\n

a + (n \u2013 1)d = -150<\/p>\n\n\n\n

11 + (n \u2013 1)(-3) = -150<\/p>\n\n\n\n

11 \u2013 3n + 3 = -150<\/p>\n\n\n\n

3n = 150 + 14<\/p>\n\n\n\n

3n = 164<\/p>\n\n\n\n

\u21d2 n = 164\/3, which is not a whole number.<\/p>\n\n\n\n

Therefore, -150 is not a term in the A.P.<\/p>\n\n\n\n

4. How many terms are there in the A.P.?<\/strong><\/p>\n\n\n\n

(i) 7, 10, 13, \u2026.., 43<\/strong><\/p>\n\n\n\n

(ii) -1, -5\/6, -2\/3, -1\/2, \u2026 , 10\/3<\/strong><\/p>\n\n\n\n

(iii) 7, 13, 19, \u2026, 205<\/strong><\/p>\n\n\n\n

(iv) 18, 15\u00bd, 13, \u2026., -47<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given, A.P. 7, 10, 13, \u2026.., 43<\/p>\n\n\n\n

Here, a = 7 and d = a2<\/sub> \u2013 a1<\/sub> = 10 \u2013 7 = 3<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

And, given nth<\/sup> term an<\/sub> = 43<\/p>\n\n\n\n

a + (n \u2013 1)d = 43<\/p>\n\n\n\n

7 + (n \u2013 1)(3) = 43<\/p>\n\n\n\n

7 + 3n \u2013 3 = 43<\/p>\n\n\n\n

3n = 43 \u2013 4<\/p>\n\n\n\n

3n = 39<\/p>\n\n\n\n

\u21d2 n = 13<\/p>\n\n\n\n

Therefore, there are 13 terms in the given A.P.<\/p>\n\n\n\n

(ii) Given, A.P. -1, -5\/6, -2\/3, -1\/2, \u2026 , 10\/3<\/p>\n\n\n\n

Here, a = -1 and d = a2<\/sub> \u2013 a1<\/sub> = -5\/6 \u2013 (-1) = 1\/6<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

And, given nth<\/sup> term an<\/sub> = 10\/3<\/p>\n\n\n\n

a + (n \u2013 1)d = 10\/3<\/p>\n\n\n\n

-1 + (n \u2013 1)(1\/6) = 10\/3<\/p>\n\n\n\n

-1 + n\/6 \u2013 1\/6 = 10\/3<\/p>\n\n\n\n

n\/6 = 10\/3 + 1 + 1\/6<\/p>\n\n\n\n

n\/6 = (20 + 6 + 1)\/6<\/p>\n\n\n\n

n = (20 + 6 + 1)<\/p>\n\n\n\n

\u21d2 n = 27<\/p>\n\n\n\n

Therefore, there are 27 terms in the given A.P.<\/p>\n\n\n\n

(iii) Given, A.P. 7, 13, 19, \u2026, 205<\/p>\n\n\n\n

Here, a = 7 and d = a2<\/sub> \u2013 a1<\/sub> = 13 \u2013 7 = 6<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

And, given nth<\/sup> term an<\/sub> = 205<\/p>\n\n\n\n

a + (n \u2013 1)d = 205<\/p>\n\n\n\n

7 + (n \u2013 1)(6) = 205<\/p>\n\n\n\n

7 + 6n \u2013 6 = 205<\/p>\n\n\n\n

6n = 205 \u2013 1<\/p>\n\n\n\n

n = 204\/6<\/p>\n\n\n\n

\u21d2 n = 34<\/p>\n\n\n\n

Therefore, there are 34 terms in the given A.P.<\/p>\n\n\n\n

(iv) Given, A.P. 18, 15\u00bd, 13, \u2026., -47<\/p>\n\n\n\n

Here, a = 7 and d = a2<\/sub> \u2013 a1<\/sub> = 15\u00bd \u2013 18 = 5\/2<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

And, given nth<\/sup> term an<\/sub> = -47<\/p>\n\n\n\n

a + (n \u2013 1)d = 43<\/p>\n\n\n\n

18 + (n \u2013 1)(-5\/2) = -47<\/p>\n\n\n\n

18 \u2013 5n\/2 + 5\/2 = -47<\/p>\n\n\n\n

36 \u2013 5n + 5 = -94<\/p>\n\n\n\n

5n = 94 + 36 + 5<\/p>\n\n\n\n

5n = 135<\/p>\n\n\n\n

\u21d2 n = 27<\/p>\n\n\n\n

Therefore, there are 27 terms in the given A.P.<\/p>\n\n\n\n

5. The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a = 5 and d = 3<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So, for the given A.P. an<\/sub> = 5 + (n \u2013 1)3 = 3n + 2<\/p>\n\n\n\n

Also given, last term = 80<\/p>\n\n\n\n

\u21d2 3n + 2 = 80<\/p>\n\n\n\n

3n = 78<\/p>\n\n\n\n

n = 78\/3 = 26<\/p>\n\n\n\n

Therefore, there are 26 terms in the A.P.<\/p>\n\n\n\n

6. The 6th<\/sup> and 17th<\/sup> terms of an A.P. are 19 and 41 respectively, find the 40th<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a6<\/sub> = 19 and a17<\/sub> = 41<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a6<\/sub> = a + (6-1)d<\/p>\n\n\n\n

\u21d2 a + 5d = 19 \u2026\u2026 (i)<\/p>\n\n\n\n

Similarity,<\/p>\n\n\n\n

a17<\/sub> = a + (17 \u2013 1)d<\/p>\n\n\n\n

\u21d2 a + 16d = 41 \u2026\u2026 (ii)<\/p>\n\n\n\n

Solving (i) and (ii),<\/p>\n\n\n\n

(ii) \u2013 (i) \u21d2<\/p>\n\n\n\n

a + 16d \u2013 (a + 5d) = 41 \u2013 19<\/p>\n\n\n\n

11d = 22<\/p>\n\n\n\n

\u21d2 d = 2<\/p>\n\n\n\n

Using d in (i), we get<\/p>\n\n\n\n

a + 5(2) = 19<\/p>\n\n\n\n

a = 19 \u2013 10 = 9<\/p>\n\n\n\n

Now, the 40th<\/sup> term is given by a40<\/sub> = 9 + (40 \u2013 1)2 = 9 + 78 = 87<\/p>\n\n\n\n

Therefore the 40th<\/sup> term is 87.<\/p>\n\n\n\n

7. If 9th<\/sup> term of an A.P. is zero, prove its 29th<\/sup> term is double the 19th<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a9<\/sub> = 0<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So, a + (9 \u2013 1)d = 0 \u21d2 a + 8d = 0 \u2026\u2026(i)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

29th<\/sup> term is given by a29<\/sub> = a + (29 \u2013 1)d<\/p>\n\n\n\n

\u21d2 a29<\/sub> = a + 28d<\/p>\n\n\n\n

And, a29<\/sub> = (a + 8d) + 20d [using (i)]<\/p>\n\n\n\n

\u21d2 a29<\/sub> = 20d \u2026.. (ii)<\/p>\n\n\n\n

Similarly, 19th<\/sup> term is given by a19<\/sub> = a + (19 \u2013 1)d<\/p>\n\n\n\n

\u21d2 a19<\/sub> = a + 18d<\/p>\n\n\n\n

And, a19<\/sub> = (a + 8d) + 10d [using (i)]<\/p>\n\n\n\n

\u21d2 a19<\/sub> = 10d \u2026..(iii)<\/p>\n\n\n\n

On comparing (ii) and (iii), it\u2019s clearly seen that<\/p>\n\n\n\n

a29<\/sub> = 2(a19<\/sub>)<\/p>\n\n\n\n

Therefore, 29th<\/sup> term is double the 19th<\/sup> term.<\/p>\n\n\n\n

8. If 10 times the 10th<\/sup> term of an A.P. is equal to 15 times the 15th<\/sup> term, show that 25th<\/sup> term of the A.P. is zero.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

10 times the 10th<\/sup> term of an A.P. is equal to 15 times the 15th<\/sup> term.<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 10(a10<\/sub>) = 15(a15<\/sub>)<\/p>\n\n\n\n

10(a + (10 \u2013 1)d) = 15(a + (15 \u2013 1)d)<\/p>\n\n\n\n

10(a + 9d) = 15(a + 14d)<\/p>\n\n\n\n

10a + 90d = 15a + 210d<\/p>\n\n\n\n

5a + 120d = 0<\/p>\n\n\n\n

5(a + 24d) = 0<\/p>\n\n\n\n

a + 24d = 0<\/p>\n\n\n\n

a + (25 \u2013 1)d = 0<\/p>\n\n\n\n

\u21d2 a25<\/sub> = 0<\/p>\n\n\n\n

Therefore, the 25th<\/sup> term of the A.P. is zero.<\/p>\n\n\n\n

9. The 10th<\/sup> and 18th<\/sup> terms of an A.P. are 41 and 73 respectively. Find 26th<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

A10<\/sub> = 41 and a18<\/sub> = 73<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a10<\/sub> = a + (10 \u2013 1)d<\/p>\n\n\n\n

\u21d2 a + 9d = 41 \u2026\u2026 (i)<\/p>\n\n\n\n

Similarity,<\/p>\n\n\n\n

a18<\/sub> = a + (18 \u2013 1)d<\/p>\n\n\n\n

\u21d2 a + 17d = 73 \u2026\u2026 (ii)<\/p>\n\n\n\n

Solving (i) and (ii),<\/p>\n\n\n\n

(ii) \u2013 (i) \u21d2<\/p>\n\n\n\n

a + 17d \u2013 (a + 9d) = 73 \u2013 41<\/p>\n\n\n\n

8d = 32<\/p>\n\n\n\n

\u21d2 d = 4<\/p>\n\n\n\n

Using d in (i), we get<\/p>\n\n\n\n

a + 9(4) = 41<\/p>\n\n\n\n

a = 41 \u2013 36 = 5<\/p>\n\n\n\n

Now, the 26th<\/sup> term is given by a26<\/sub> = 5 + (26 \u2013 1)4 = 5 + 100 = 105<\/p>\n\n\n\n

Therefore the 26th<\/sup> term is 105.<\/p>\n\n\n\n

10. In a certain A.P. the 24th<\/sup> term is twice the 10th<\/sup> term. Prove that the 72nd<\/sup> term is twice the 34th<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

24th<\/sup> term is twice the 10th<\/sup> term.<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 a24<\/sub> = 2(a10<\/sub>)<\/p>\n\n\n\n

a + (24 \u2013 1)d = 2(a + (10 \u2013 1)d)<\/p>\n\n\n\n

a + 23d = 2(a + 9d)<\/p>\n\n\n\n

a + 23d = 2a + 18d<\/p>\n\n\n\n

a = 5d \u2026. (1)<\/p>\n\n\n\n

Now, the 72nd<\/sup> term can be expressed as<\/p>\n\n\n\n

a72<\/sub> = a + (72 \u2013 1)d<\/p>\n\n\n\n

= a + 71d<\/p>\n\n\n\n

= a + 5d + 66d<\/p>\n\n\n\n

= a + a + 66d [using (1)]<\/p>\n\n\n\n

= 2(a + 33d)<\/p>\n\n\n\n

= 2(a + (34 \u2013 1)d)<\/p>\n\n\n\n

= 2(a34<\/sub>)<\/p>\n\n\n\n

\u21d2 a72 <\/sub>= 2(a34<\/sub>)<\/p>\n\n\n\n

Hence, the 72nd<\/sup> term is twice the 34th<\/sup> term of the given A.P.<\/p>\n\n\n\n

11. The 26th<\/sup>, 11th<\/sup> and the last term of an A.P. are 0, 3 and -1\/5, respectively. Find the common difference and the number of terms.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a26 <\/sub>= 0 , a11 <\/sub>= 3 and an <\/sub>(last term) = -1\/5 of an A.P.<\/p>\n\n\n\n

We know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

a26<\/sub> = a + (26 \u2013 1)d<\/p>\n\n\n\n

\u21d2 a + 25d = 0 \u2026..(1)<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

a11<\/sub> = a + (11 \u2013 1)d<\/p>\n\n\n\n

\u21d2 a + 10d = 3 \u2026\u2026 (2)<\/p>\n\n\n\n

Solving (1) and (2),<\/p>\n\n\n\n

(1) \u2013 (2) \u21d2<\/p>\n\n\n\n

a + 25d \u2013 (a + 10d) = 0 \u2013 3<\/p>\n\n\n\n

15d = -3<\/p>\n\n\n\n

\u21d2 d = -1\/5<\/p>\n\n\n\n

Using d in (1), we get<\/p>\n\n\n\n

a + 25(-1\/5) = 0<\/p>\n\n\n\n

a = 5<\/p>\n\n\n\n

Now, given that the last term an<\/sub> = -1\/5<\/p>\n\n\n\n

\u21d2 5 + (n \u2013 1)(-1\/5) = -1\/5<\/p>\n\n\n\n

5 + -n\/5 + 1\/5 = -1\/5<\/p>\n\n\n\n

25 \u2013 n + 1 = -1<\/p>\n\n\n\n

n = 27<\/p>\n\n\n\n

Therefore, the A.P has 27 terms and its common difference is -1\/5.<\/p>\n\n\n\n

12. If the nth<\/sup> term of the A.P. 9, 7, 5, \u2026. is same as the nth<\/sup> term of the A.P. 15, 12, 9, \u2026 find n.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

A.P1 <\/sub>= 9, 7, 5, \u2026. and A.P2 <\/sub>= 15, 12, 9, \u2026<\/p>\n\n\n\n

And, we know that, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

For A.P1<\/sub>,<\/p>\n\n\n\n

a = 9, d = Second term \u2013 first term = 9 \u2013 7 = -2<\/p>\n\n\n\n

And, its nth<\/sup> term an <\/sub>= 9 + (n \u2013 1)(-2) = 9 \u2013 2n + 2<\/p>\n\n\n\n

an <\/sub>= 11 \u2013 2n \u2026..(i)<\/p>\n\n\n\n

Similarly, for A.P2<\/sub><\/p>\n\n\n\n

a = 15, d = Second term \u2013 first term = 12 \u2013 15 = -3<\/p>\n\n\n\n

And, its nth<\/sup> term an <\/sub>= 15 + (n \u2013 1)(-3) = 15 \u2013 3n + 3<\/p>\n\n\n\n

an <\/sub>= 18 \u2013 3n \u2026..(ii)<\/p>\n\n\n\n

According to the question, its given that<\/p>\n\n\n\n

nth<\/sup> term of the A.P1<\/sub> = nth<\/sup> term of the A.P2<\/sub><\/p>\n\n\n\n

\u21d2 11 \u2013 2n = 18 \u2013 3n<\/p>\n\n\n\n

n = 7<\/p>\n\n\n\n

Therefore, the 7th<\/sup> term of the both the A.Ps are equal.<\/p>\n\n\n\n

13. Find the 12th<\/sup> term from the end of the following arithmetic progressions:<\/strong><\/p>\n\n\n\n

(i) 3, 5, 7, 9, \u2026. 201<\/strong><\/p>\n\n\n\n

(ii) 3,8,13, \u2026 ,253<\/strong><\/p>\n\n\n\n

(iii) 1, 4, 7, 10, \u2026 ,88<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In order the find the 12th<\/sup> term for the end of an A.P. which has n terms, its done by simply finding the ((n -12) + 1)th<\/sup> of the A.P<\/p>\n\n\n\n

And we know, nth<\/sup> term an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

(i) Given A.P = 3, 5, 7, 9, \u2026. 201<\/p>\n\n\n\n

Here, a = 3 and d = (5 \u2013 3) = 2<\/p>\n\n\n\n

Now, find the number of terms when the last term is known i.e, 201<\/p>\n\n\n\n

an<\/sub> = 3 + (n \u2013 1)2 = 201<\/p>\n\n\n\n

3 + 2n \u2013 2 = 201<\/p>\n\n\n\n

2n = 200<\/p>\n\n\n\n

n = 100<\/p>\n\n\n\n

Hence, the A.P has 100 terms.<\/p>\n\n\n\n

So, the 12th<\/sup> term from the end is same as (100 \u2013 12 + 1)th<\/sup> of the A.P which is the 89th<\/sup> term.<\/p>\n\n\n\n

\u21d2 a89 <\/sub>= 3 + (89 \u2013 1)2<\/p>\n\n\n\n

= 3 + 88(2)<\/p>\n\n\n\n

= 3 + 176<\/p>\n\n\n\n

= 179<\/p>\n\n\n\n

Therefore, the 12th<\/sup> term from the end of the A.P is 179.<\/p>\n\n\n\n

(ii) Given A.P = 3,8,13, \u2026 ,253<\/p>\n\n\n\n

Here, a = 3 and d = (8 \u2013 3) = 5<\/p>\n\n\n\n

Now, find the number of terms when the last term is known i.e, 253<\/p>\n\n\n\n

an<\/sub> = 3 + (n \u2013 1)5 = 253<\/p>\n\n\n\n

3 + 5n \u2013 5 = 253<\/p>\n\n\n\n

5n = 253 + 2 = 255<\/p>\n\n\n\n

n = 255\/5<\/p>\n\n\n\n

n = 51<\/p>\n\n\n\n

Hence, the A.P has 51 terms.<\/p>\n\n\n\n

So, the 12th<\/sup> term from the end is same as (51 \u2013 12 + 1)th<\/sup> of the A.P which is the 40th<\/sup> term.<\/p>\n\n\n\n

\u21d2 a40 <\/sub>= 3 + (40 \u2013 1)5<\/p>\n\n\n\n

= 3 + 39(5)<\/p>\n\n\n\n

= 3 + 195<\/p>\n\n\n\n

= 198<\/p>\n\n\n\n

Therefore, the 12th<\/sup> term from the end of the A.P is 198.<\/p>\n\n\n\n

(iii) Given A.P = 1, 4, 7, 10, \u2026 ,88<\/p>\n\n\n\n

Here, a = 1 and d = (4 \u2013 1) = 3<\/p>\n\n\n\n

Now, find the number of terms when the last term is known i.e, 88<\/p>\n\n\n\n

an<\/sub> = 1 + (n \u2013 1)3 = 88<\/p>\n\n\n\n

1 + 3n \u2013 3 = 88<\/p>\n\n\n\n

3n = 90<\/p>\n\n\n\n

n = 30<\/p>\n\n\n\n

Hence, the A.P has 30 terms.<\/p>\n\n\n\n

So, the 12th<\/sup> term from the end is same as (30 \u2013 12 + 1)th<\/sup> of the A.P which is the 19th<\/sup> term.<\/p>\n\n\n\n

\u21d2 a89 <\/sub>= 1 + (19 \u2013 1)3<\/p>\n\n\n\n

= 1 + 18(3)<\/p>\n\n\n\n

= 1 + 54<\/p>\n\n\n\n

= 55<\/p>\n\n\n\n

Therefore, the 12th<\/sup> term from the end of the A.P is 55.<\/p>\n\n\n\n

14. The 4th<\/sup> term of an A.P. is three times the first and the 7th<\/sup> term exceeds twice the third term by 1. Find the first term and the common difference.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s consider the first term and the common difference of the A.P to be a and d respectively.<\/p>\n\n\n\n

Then, we know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Given conditions,<\/p>\n\n\n\n

4th<\/sup> term of an A.P. is three times the first<\/p>\n\n\n\n

Expressing this by equation we have,<\/p>\n\n\n\n

\u21d2 a4<\/sub> = 3(a)<\/p>\n\n\n\n

a + (4 \u2013 1)d = 3a<\/p>\n\n\n\n

3d = 2a \u21d2 a = 3d\/2\u2026\u2026.(i)<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

7th<\/sup> term exceeds twice the third term by 1<\/p>\n\n\n\n

\u21d2 a7<\/sub> = 2(a3<\/sub>) + 1<\/p>\n\n\n\n

a + (7 \u2013 1)d = 2(a + (3\u20131)d) + 1<\/p>\n\n\n\n

a + 6d = 2a + 4d + 1<\/p>\n\n\n\n

a \u2013 2d +1 = 0 \u2026.. (ii)<\/p>\n\n\n\n

Using (i) in (ii), we have<\/p>\n\n\n\n

3d\/2 \u2013 2d + 1 = 0<\/p>\n\n\n\n

3d \u2013 4d + 2 = 0<\/p>\n\n\n\n

d = 2<\/p>\n\n\n\n

So, putting d = 2 in (i), we get a<\/p>\n\n\n\n

\u21d2 a = 3<\/p>\n\n\n\n

Therefore, the first term is 3 and the common difference is 2.<\/p>\n\n\n\n

15. Find the second term and the nth<\/sup> term of an A.P. whose 6th<\/sup> term is 12 and the 8th<\/sup> term is 22.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, in an A.P<\/p>\n\n\n\n

a6<\/sub> = 12 and a8<\/sub> = 22<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a6 <\/sub>= a + (6-1)d = a + 5d = 12 \u2026. (i)<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

a8<\/sub> = a + (8-1)d = a + 7d = 22 \u2026\u2026. (ii)<\/p>\n\n\n\n

Solving (i) and (ii), we have<\/p>\n\n\n\n

(ii) \u2013 (i) \u21d2<\/p>\n\n\n\n

a + 7d \u2013 (a + 5d) = 22 \u2013 12<\/p>\n\n\n\n

2d = 10<\/p>\n\n\n\n

d = 5<\/p>\n\n\n\n

Putting d in (i) we get,<\/p>\n\n\n\n

a + 5(5) = 12<\/p>\n\n\n\n

a = 12 \u2013 25<\/p>\n\n\n\n

a = -13<\/p>\n\n\n\n

Thus, for the A.P: a = -13 and d = 5<\/p>\n\n\n\n

So, the nth<\/sup> term is given by an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

an <\/sub>= -13 + (n-1)5 = -13 + 5n \u2013 5<\/p>\n\n\n\n

\u21d2 an <\/sub>= 5n \u2013 18<\/p>\n\n\n\n

Hence, the second term is given by a2 <\/sub>= 5(2) \u2013 18 = 10 \u2013 18 = -8<\/p>\n\n\n\n

16. How many numbers of two digit are divisible by 3?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The first 2 digit number divisible by 3 is 12. And, the last 2 digit number divisible by 3 is 99.<\/p>\n\n\n\n

So, this forms an A.P.<\/p>\n\n\n\n

12, 15, 18, 21, \u2026. , 99<\/p>\n\n\n\n

Where, a = 12 and d = 3<\/p>\n\n\n\n

Finding the number of terms in this A.P<\/p>\n\n\n\n

\u21d2 99 = 12 + (n-1)3<\/p>\n\n\n\n

99 = 12 + 3n \u2013 3<\/p>\n\n\n\n

90 = 3n<\/p>\n\n\n\n

n = 90\/3 = 30<\/p>\n\n\n\n

Therefore, there are 30 two digit numbers divisible by 3.<\/p>\n\n\n\n

17. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, an A.P of 60 terms<\/p>\n\n\n\n

And, a = 7 and a60<\/sub> = 125<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 a60<\/sub> = 7 + (60 \u2013 1)d = 125<\/p>\n\n\n\n

7 + 59d = 125<\/p>\n\n\n\n

59d = 118<\/p>\n\n\n\n

d = 2<\/p>\n\n\n\n

So, the 32nd<\/sup> term is given by<\/p>\n\n\n\n

a32<\/sub> = 7 + (32 -1)2 = 7 + 62 = 69<\/p>\n\n\n\n

\u21d2 a32<\/sub> = 69<\/p>\n\n\n\n

18. The sum of 4th<\/sup> and 8th<\/sup> terms of an A.P. is 24 and the sum of the 6th<\/sup> and 10th<\/sup> terms is 34. Find the first term and the common difference of the A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, in an A.P<\/p>\n\n\n\n

The sum of 4th<\/sup> and 8th<\/sup> terms of an A.P. is 24<\/p>\n\n\n\n

\u21d2 a4<\/sub> + a8<\/sub> = 24<\/p>\n\n\n\n

And, we know that an<\/sub> = a + (n \u2013 1)d[a + (4-1)d] + [a + (8-1)d] = 24<\/p>\n\n\n\n

2a + 10d = 24<\/p>\n\n\n\n

a + 5d = 12 \u2026. (i)<\/p>\n\n\n\n

Also given that,<\/p>\n\n\n\n

the sum of the 6th<\/sup> and 10th<\/sup> terms is 34<\/p>\n\n\n\n

\u21d2 a6<\/sub> + a10<\/sub> = 34[a + 5d] + [a + 9d] = 34<\/p>\n\n\n\n

2a + 14d = 34<\/p>\n\n\n\n

a + 7d = 17 \u2026\u2026 (ii)<\/p>\n\n\n\n

Subtracting (i) form (ii), we have<\/p>\n\n\n\n

a + 7d \u2013 (a + 5d) = 17 \u2013 12<\/p>\n\n\n\n

2d = 5<\/p>\n\n\n\n

d = 5\/2<\/p>\n\n\n\n

Using d in (i) we get,<\/p>\n\n\n\n

a + 5(5\/2) = 12<\/p>\n\n\n\n

a = 12 \u2013 25\/2<\/p>\n\n\n\n

a = -1\/2<\/p>\n\n\n\n

Therefore, the first term is -1\/2 and the common difference is 5\/2.<\/p>\n\n\n\n

19. The first term of an A.P. is 5 and its 100th<\/sup> term is -292. Find the 50th<\/sup> term of this A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, an A.P whose<\/p>\n\n\n\n

a = 5 and a100<\/sub> = -292<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

a100 = 5 + 99d = -292<\/p>\n\n\n\n

99d = -297<\/p>\n\n\n\n

d = -3<\/p>\n\n\n\n

Hence, the 50th<\/sup> term is<\/p>\n\n\n\n

a50<\/sub> = a + 49d = 5 + 49(-3) = 5 \u2013 147 = -142<\/p>\n\n\n\n

20. Find a30<\/sub> \u2013 a20<\/sub> for the A.P.<\/strong><\/p>\n\n\n\n

(i) -9, -14, -19, -24 (ii) a, a+d, a+2d, a+3d, \u2026\u2026<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So, a30<\/sub> \u2013 a20<\/sub> = (a + 29d) \u2013 (a + 19) =10d<\/p>\n\n\n\n

(i) Given A.P. -9, -14, -19, -24<\/p>\n\n\n\n

Here, a = -9 and d = -14 \u2013 (-9) = = -14 + 9 = -5<\/p>\n\n\n\n

So, a30<\/sub> \u2013 a20<\/sub> = 10d<\/p>\n\n\n\n

= 10(-5)<\/p>\n\n\n\n

= -50<\/p>\n\n\n\n

(ii) Given A.P. a, a+d, a+2d, a+3d, \u2026\u2026<\/p>\n\n\n\n

So, a30<\/sub> \u2013 a20<\/sub> = (a + 29d) \u2013 (a + 19d)<\/p>\n\n\n\n

=10d<\/p>\n\n\n\n

21. Write the expression an<\/sub> \u2013 ak<\/sub> for the A.P. a, a+d, a+2d, \u2026..<\/strong><\/p>\n\n\n\n

Hence, find the common difference of the A.P. for which<\/strong><\/p>\n\n\n\n

(i) 11th<\/sup> term is 5 and 13th<\/sup> term is 79.<\/strong><\/p>\n\n\n\n

(ii) a10 <\/sub>\u2013 a5<\/sub> = 200<\/strong><\/p>\n\n\n\n

(iii) 20th<\/sup> term is 10 more than the 18th<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given A.P. a, a+d, a+2d, \u2026..<\/p>\n\n\n\n

So, an <\/sub>= a + (n-1)d = a + nd \u2013d<\/p>\n\n\n\n

And, ak <\/sub>= a + (k-1)d = a + kd \u2013 d<\/p>\n\n\n\n

an <\/sub>\u2013 ak <\/sub>= (a + nd \u2013 d) \u2013 (a + kd \u2013 d)<\/p>\n\n\n\n

= (n \u2013 k)d<\/p>\n\n\n\n

(i) Given 11th<\/sup> term is 5 and 13th<\/sup> term is 79,<\/p>\n\n\n\n

Here n = 13 and k = 11,<\/p>\n\n\n\n

a13 <\/sub>\u2013 a11 <\/sub>= (13 \u2013 11)d = 2d<\/p>\n\n\n\n

\u21d2 79 \u2013 5 = 2d<\/p>\n\n\n\n

d = 74\/2 = 37<\/p>\n\n\n\n

(ii) Given, a10 <\/sub>\u2013 a5<\/sub> = 200<\/p>\n\n\n\n

\u21d2 (10 \u2013 5)d = 200<\/p>\n\n\n\n

5d = 200<\/p>\n\n\n\n

d = 40<\/p>\n\n\n\n

(iii) Given, 20th<\/sup> term is 10 more than the 18th<\/sup> term.<\/p>\n\n\n\n

\u21d2 a20<\/sub> \u2013 a18<\/sub> = 10<\/p>\n\n\n\n

(20 \u2013 18)d = 10<\/p>\n\n\n\n

2d = 10<\/p>\n\n\n\n

d = 5<\/p>\n\n\n\n

22. Find n if the given value of x is the nth<\/sup> term of the given A.P.<\/strong><\/p>\n\n\n\n

(i) 25, 50, 75, 100, ; x = 1000 (ii) -1, -3, -5, -7, \u2026; x = -151<\/strong><\/p>\n\n\n\n

(iii) 5\u00bd, 11, 16\u00bd, 22, \u2026.; x = 550 (iv) 1, 21\/11, 31\/11, 41\/11, \u2026; x = 171\/11<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given A.P. 25, 50, 75, 100, \u2026\u2026 ,1000<\/p>\n\n\n\n

Here, a = 25 d = 50 \u2013 25 = 25<\/p>\n\n\n\n

Last term (nth<\/sup> term) = 1000<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 1000 = 25 + (n-1)25<\/p>\n\n\n\n

1000 = 25 + 25n \u2013 25<\/p>\n\n\n\n

n = 1000\/25<\/p>\n\n\n\n

n = 40<\/p>\n\n\n\n

(ii) Given A.P. -1, -3, -5, -7, \u2026., -151<\/p>\n\n\n\n

Here, a = -1 d = -3 \u2013 (-1) = -2<\/p>\n\n\n\n

Last term (nth<\/sup> term) = -151<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 -151 = -1 + (n-1)(-2)<\/p>\n\n\n\n

-151 = -1 \u2013 2n + 2<\/p>\n\n\n\n

n = 152\/2<\/p>\n\n\n\n

n = 76<\/p>\n\n\n\n

(iii) Given A.P. 5\u00bd, 11, 16\u00bd, 22, \u2026 , 550<\/p>\n\n\n\n

Here, a = 5\u00bd d = 11 \u2013 (5\u00bd) = 5\u00bd = 11\/2<\/p>\n\n\n\n

Last term (nth<\/sup> term) = 550<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 550 = 5\u00bd + (n-1)(11\/2)<\/p>\n\n\n\n

550 x 2 = 11+ 11n \u2013 11<\/p>\n\n\n\n

1100 = 11n<\/p>\n\n\n\n

n = 100<\/p>\n\n\n\n

(iv) Given A.P. 1, 21\/11, 31\/11, 41\/11, 171\/11<\/p>\n\n\n\n

Here, a = 1 d = 21\/11 \u2013 1 = 10\/11<\/p>\n\n\n\n

Last term (nth<\/sup> term) = 171\/11<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 171\/11 = 1 + (n-1)10\/11<\/p>\n\n\n\n

171 = 11 + 10n \u2013 10<\/p>\n\n\n\n

n = 170\/10<\/p>\n\n\n\n

n = 17<\/p>\n\n\n\n

23. The eighth term of an A.P is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, an A.P in which,<\/p>\n\n\n\n

a8<\/sub> = 1\/2(a2<\/sub>)<\/p>\n\n\n\n

a11<\/sub> = 1\/3(a4<\/sub>) + 1<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 a8<\/sub> = 1\/2(a2<\/sub>)<\/p>\n\n\n\n

a + 7d = 1\/2(a + d)<\/p>\n\n\n\n

2a + 14d = a + d<\/p>\n\n\n\n

a + 13d = 0 \u2026\u2026 (i)<\/p>\n\n\n\n

And, a11<\/sub> = 1\/3(a4<\/sub>) + 1<\/p>\n\n\n\n

a + 10d = 1\/3(a + 3d) + 1<\/p>\n\n\n\n

3a + 30d = a + 3d + 3<\/p>\n\n\n\n

2a + 27d = 3 \u2026\u2026 (ii)<\/p>\n\n\n\n

Solving (i) and (ii), by (ii) \u2013 2x(i) \u21d2<\/p>\n\n\n\n

2a + 27d \u2013 2(a + 13d) = 3 \u2013 0<\/p>\n\n\n\n

d = 3<\/p>\n\n\n\n

Putting d in (i) we get,<\/p>\n\n\n\n

a + 13(3) = 0<\/p>\n\n\n\n

a = -39<\/p>\n\n\n\n

Thus, the 15th<\/sup> term a15 <\/sub>= -39 + 14(3) = -39 + 42 = 3<\/p>\n\n\n\n

24. Find the arithmetic progression whose third term is 16 and the seventh term exceeds its fifth term by 12.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, in an A.P<\/p>\n\n\n\n

a3<\/sub> = 16 and a7<\/sub> = a5<\/sub> + 12<\/p>\n\n\n\n

We know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u21d2 a + 2d = 16\u2026\u2026 (i)<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

a + 6d = a + 4d + 12<\/p>\n\n\n\n

2d = 12<\/p>\n\n\n\n

\u21d2 d = 6<\/p>\n\n\n\n

Using d in (i), we have<\/p>\n\n\n\n

a + 2(6) = 16<\/p>\n\n\n\n

a = 16 \u2013 12 = 4<\/p>\n\n\n\n

Hence, the A.P is 4, 10, 16, 22, \u2026\u2026.<\/p>\n\n\n\n

25. The 7th<\/sup> term of an A.P. is 32 and its 13th<\/sup> term is 62. Find the A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a7 <\/sub>= 32 and a13 <\/sub>= 62<\/p>\n\n\n\n

From an <\/sub>\u2013 ak <\/sub>= (a + nd \u2013 d) \u2013 (a + kd \u2013 d)<\/p>\n\n\n\n

= (n \u2013 k)d<\/p>\n\n\n\n

a13<\/sub> \u2013 a7<\/sub> = (13 \u2013 7)d = 62 \u2013 32 = 30<\/p>\n\n\n\n

6d = 30<\/p>\n\n\n\n

d = 5<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

a7<\/sub> = a + (7 \u2013 1)5 = 32<\/p>\n\n\n\n

a + 30 = 32<\/p>\n\n\n\n

a = 2<\/p>\n\n\n\n

Hence, the A.P is 2, 7, 12, 17, \u2026\u2026<\/p>\n\n\n\n

26. Which term of the A.P. 3, 10, 17, \u2026. will be 84 more than its 13th<\/sup> term ?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, A.P. 3, 10, 17, \u2026.<\/p>\n\n\n\n

Here, a = 3 and d = 10 \u2013 3 = 7<\/p>\n\n\n\n

According the question,<\/p>\n\n\n\n

an<\/sub> = a13<\/sub> + 84<\/p>\n\n\n\n

Using an<\/sub> = a + (n \u2013 1)d,<\/p>\n\n\n\n

3 + (n \u2013 1)7 = 3 + (13 \u2013 1)7 + 84<\/p>\n\n\n\n

3 + 7n \u2013 7 = 3 + 84 + 84<\/p>\n\n\n\n

7n = 168 + 7<\/p>\n\n\n\n

n = 175\/7<\/p>\n\n\n\n

n = 25<\/p>\n\n\n\n

Therefore, it the 25th<\/sup> term which is 84 more than its 13th<\/sup> term.<\/p>\n\n\n\n

27. Two arithmetic progressions have the same common difference. The difference between their 100th<\/sup> terms is 100, what is the difference between their 1000th<\/sup> terms?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the two A.Ps be A.P1<\/sub> and A.P2<\/sub><\/p>\n\n\n\n

For A.P1<\/sub> the first term = a and the common difference = d<\/p>\n\n\n\n

And for A.P2<\/sub> the first term = b and the common difference = d<\/p>\n\n\n\n

So, from the question we have<\/p>\n\n\n\n

a100<\/sub> \u2013 b100<\/sub> = 100<\/p>\n\n\n\n

(a + 99d) \u2013 (b + 99d) = 100<\/p>\n\n\n\n

a \u2013 b = 100<\/p>\n\n\n\n

Now, the difference between their 1000th<\/sup> terms is,<\/p>\n\n\n\n

(a + 999d) \u2013 (b + 999d) = a \u2013 b = 100<\/p>\n\n\n\n

Therefore, the difference between their 1000th<\/sup> terms is also 100.<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.5 Page No: 9.30<\/h4>\n\n\n\n

1. Find the value of x for which (8x + 4), (6x \u2013 2) and (2x + 7) are in A.P.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

(8x + 4), (6x \u2013 2) and (2x + 7) are in A.P.<\/p>\n\n\n\n

So, the common difference between the consecutive terms should be the same.<\/p>\n\n\n\n

(6x \u2013 2) \u2013 (8x + 4) = (2x + 7) \u2013 (6x \u2013 2)<\/p>\n\n\n\n

\u21d2 6x \u2013 2 \u2013 8x \u2013 4 = 2x + 7 \u2013 6x + 2<\/p>\n\n\n\n

\u21d2 -2x \u2013 6 = -4x + 9<\/p>\n\n\n\n

\u21d2 -2x + 4x = 9 + 6<\/p>\n\n\n\n

\u21d2 2x = 15<\/p>\n\n\n\n

Therefore, x = 15\/2<\/p>\n\n\n\n

2. If x + 1, 3x and 4x + 2 are in A.P., find the value of x.
Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

x + 1, 3x and 4x + 2 are in A.P.<\/p>\n\n\n\n

So, the common difference between the consecutive terms should be the same.<\/p>\n\n\n\n

3x \u2013 x \u2013 1 = 4x + 2 \u2013 3x<\/p>\n\n\n\n

\u21d2 2x \u2013 1 = x + 2<\/p>\n\n\n\n

\u21d2 2x \u2013 x = 2 + 1<\/p>\n\n\n\n

\u21d2 x = 3<\/p>\n\n\n\n

Therefore, x = 3<\/p>\n\n\n\n

3. Show that (a \u2013 b)\u00b2, (a\u00b2 + b\u00b2) and (a + b)\u00b2 are in A.P.
Solution:<\/strong><\/p>\n\n\n\n

If (a \u2013 b)\u00b2, (a\u00b2 + b\u00b2) and (a + b)\u00b2 have to be in A.P. then,<\/p>\n\n\n\n

It should satisfy the condition,<\/p>\n\n\n\n

2b = a + c [for a, b, c are in A.P]<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

2 (a\u00b2 + b\u00b2) = (a \u2013 b)\u00b2 + (a + b)\u00b2<\/p>\n\n\n\n

2 (a\u00b2 + b\u00b2) = a\u00b2 + b\u00b2 \u2013 2ab + a\u00b2 + b\u00b2 + 2ab<\/p>\n\n\n\n

2 (a\u00b2 + b\u00b2) = 2a\u00b2 + 2b\u00b2 = 2 (a\u00b2 + b\u00b2)<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

4. The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s consider the three terms of the A.P. to be a \u2013 d, a, a + d<\/p>\n\n\n\n

so, the sum of three terms = 21<\/p>\n\n\n\n

\u21d2 a \u2013 d + a + a + d = 21<\/p>\n\n\n\n

\u21d2 3a = 21<\/p>\n\n\n\n

\u21d2 a = 7<\/p>\n\n\n\n

And, product of the first and 3rd = 2nd term + 6<\/p>\n\n\n\n

\u21d2 (a \u2013 d) (a + d) = a + 6<\/p>\n\n\n\n

a\u00b2 \u2013 d\u00b2 = a + 6<\/p>\n\n\n\n

\u21d2 (7 )\u00b2 \u2013 d\u00b2 = 7 + 6<\/p>\n\n\n\n

\u21d2 49 \u2013 d\u00b2 = 13<\/p>\n\n\n\n

\u21d2 d\u00b2 = 49 \u2013 13 = 36<\/p>\n\n\n\n

\u21d2 d\u00b2 = (6)\u00b2<\/p>\n\n\n\n

\u21d2 d = 6<\/p>\n\n\n\n

Hence, the terms are 7 \u2013 6, 7, 7 + 6 \u21d2 1, 7, 13<\/p>\n\n\n\n

5. Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.
Solution:<\/strong><\/p>\n\n\n\n

Let the three numbers of the A.P. be a \u2013 d, a, a + d<\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Sum of these numbers = 27<\/p>\n\n\n\n

a \u2013 d + a + a + d = 27<\/p>\n\n\n\n

\u21d2 3a = 27<\/p>\n\n\n\n

a = 27\/3 = 9<\/p>\n\n\n\n

Now, product of these numbers = 648<\/p>\n\n\n\n

(a \u2013 d)(a)(a + d) = 648<\/p>\n\n\n\n

a(a2<\/sup> \u2013 d2<\/sup>) = 648<\/p>\n\n\n\n

a2<\/sup> \u2013 648\/a = d2<\/sup><\/p>\n\n\n\n

92<\/sup> \u2013 (648\/9) = d2<\/sup><\/p>\n\n\n\n

93<\/sup> \u2013 648 = 9d2<\/sup><\/p>\n\n\n\n

729 \u2013 648 = 9d2<\/sup><\/p>\n\n\n\n

81 = 9d2<\/sup><\/p>\n\n\n\n

d2<\/sup> = 9<\/p>\n\n\n\n

d = 3 or -3<\/p>\n\n\n\n

Hence, the terms are 9-3, 9 and 9+3 \u21d2 6, 9, 12 or 12, 9, 6 (for d = -3)<\/p>\n\n\n\n

6. Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s consider the four terms of the A.P. to be (a \u2013 3d), (a \u2013 d), (a + d) and (a + 3d).<\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Sum of these terms = 50<\/p>\n\n\n\n

\u21d2 (a \u2013 3d) + (a \u2013 d) + (a + d) + (a + 3d) = 50<\/p>\n\n\n\n

\u21d2 a \u2013 3d + a \u2013 d + a + d + a \u2013 3d= 50<\/p>\n\n\n\n

\u21d2 4a = 50<\/p>\n\n\n\n

\u21d2 a = 50\/4 = 25\/2<\/p>\n\n\n\n

And, also given that the greatest number = 4 x least number<\/p>\n\n\n\n

\u21d2 a + 3d = 4 (a \u2013 3d)<\/p>\n\n\n\n

\u21d2 a + 3d = 4a \u2013 12d<\/p>\n\n\n\n

\u21d2 4a \u2013 a = 3d + 12d<\/p>\n\n\n\n

\u21d23a = 15d<\/p>\n\n\n\n

\u21d2a = 5d<\/p>\n\n\n\n

Using the value of a in the above equation, we have<\/p>\n\n\n\n

\u21d225\/2 = 5d<\/p>\n\n\n\n

\u21d2 d = 5\/2<\/p>\n\n\n\n

So, the terms will be:<\/p>\n\n\n\n

(a \u2013 3d) = (25\/2 \u2013 3(5\/2)), (a \u2013 d) = (25\/2 \u2013 5\/2), (25\/2 + 5\/2) and (25\/2 + 3(5\/2)).<\/p>\n\n\n\n

\u21d2 5, 10, 15, 20<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.6 Page No: 9.50<\/h4>\n\n\n\n

1. Find the sum of the following arithmetic progressions:<\/strong><\/p>\n\n\n\n

(i) 50, 46, 42, \u2026 to 10 terms<\/strong><\/p>\n\n\n\n

(ii) 1, 3, 5, 7, \u2026 to 12 terms<\/strong><\/p>\n\n\n\n

(iii) 3, 9\/2, 6, 15\/2, \u2026 to 25 terms<\/strong><\/p>\n\n\n\n

(iv) 41, 36, 31, \u2026 to 12 terms<\/strong><\/p>\n\n\n\n

(v) a + b, a \u2013 b, a \u2013 3b, \u2026 to 22 terms<\/strong><\/p>\n\n\n\n

(vi) (x \u2013 y)2<\/sup>, (x2<\/sup> + y2<\/sup>), (x + y)2<\/sup>, to 22 tams<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(viii) \u2013 26, \u2013 24, \u2013 22, \u2026. to 36 terms<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In an A.P if the first term = a, common difference = d, and if there are n terms.<\/p>\n\n\n\n

Then, sum of n terms is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(i) Given A.P.is 50, 46, 42 to 10 term.<\/p>\n\n\n\n

First term (a) = 50<\/p>\n\n\n\n

Common difference (d) = 46 \u2013 50 = \u2013 4<\/p>\n\n\n\n

nth <\/sup>term (n) = 10<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 5{100 \u2013 9.4}<\/p>\n\n\n\n

= 5{100 \u2013 36}<\/p>\n\n\n\n

= 5 \u00d7 64<\/p>\n\n\n\n

\u2234 S10<\/sub> = 320<\/p>\n\n\n\n

(ii) Given A.P is, 1, 3, 5, 7, \u2026..to 12 terms.<\/p>\n\n\n\n

First term (a) = 1<\/p>\n\n\n\n

Common difference (d) = 3 \u2013 1 = 2<\/p>\n\n\n\n

nth<\/sup> term (n) = 12<\/p>\n\n\n\n

\"R<\/figure>\n\n\n\n

= 6 \u00d7 {2 + 22} = 6.24<\/p>\n\n\n\n

\u2234 S12<\/sub> = 144<\/p>\n\n\n\n

(iii) Given A.P. is 3, 9\/2, 6, 15\/2, \u2026 to 25 terms<\/p>\n\n\n\n

First term (a) = 3<\/p>\n\n\n\n

Common difference (d) = 9\/2 \u2013 3 = 3\/2<\/p>\n\n\n\n

Sum of n terms Sn<\/sub>, given n = 25<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv) Given expression is 41, 36, 31, \u2026.. to 12 terms.<\/p>\n\n\n\n

First term (a) = 41<\/p>\n\n\n\n

Common difference (d) = 36 \u2013 41 = -5<\/p>\n\n\n\n

Sum of n terms Sn<\/sub>, given n = 12<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(v) a + b, a \u2013 b, a \u2013 3b, \u2026.. to 22 terms<\/p>\n\n\n\n

First term (a) = a + b<\/p>\n\n\n\n

Common difference (d) = a \u2013 b \u2013 a \u2013 b = -2b<\/p>\n\n\n\n

Sum of n terms Sn<\/sub> = n\/2{2a(n \u2013 1). d}<\/p>\n\n\n\n

Here n = 22<\/p>\n\n\n\n

S22<\/sub> = 22\/2{2.(a + b) + (22 \u2013 1). -2b}<\/p>\n\n\n\n

= 11{2(a + b) \u2013 22b)<\/p>\n\n\n\n

= 11{2a \u2013 20b}<\/p>\n\n\n\n

= 22a \u2013 440b<\/p>\n\n\n\n

\u2234S22<\/sub> = 22a \u2013 440b<\/p>\n\n\n\n

(vi) (x \u2013 y)2<\/sup>,(x2<\/sup> + y2<\/sup>), (x + y)2<\/sup>,\u2026 to n terms<\/p>\n\n\n\n

First term (a) = (x \u2013 y)2<\/sup><\/p>\n\n\n\n

Common difference (d) = x2<\/sup> + y2<\/sup> \u2013 (x \u2013 y)2<\/sup><\/p>\n\n\n\n

= x2<\/sup> + y2<\/sup> \u2013 (x2<\/sup> + y2<\/sup> \u2013 2xy)<\/p>\n\n\n\n

= x2<\/sup> + y2<\/sup> \u2013 x2<\/sup> + y2<\/sup> + 2xy<\/p>\n\n\n\n

= 2xy<\/p>\n\n\n\n

Sum of nth<\/sup> terms Sn<\/sub> = n\/2{2a + (n \u2013 1). d}<\/p>\n\n\n\n

= n\/2{2(x \u2013 y)2<\/sup> + (n \u2013 1). 2xy}<\/p>\n\n\n\n

= n{(x \u2013 y)2<\/sup> + (n \u2013 1)xy}<\/p>\n\n\n\n

\u2234 Sn<\/sub> = n{(x \u2014 y)2<\/sup> + (n \u2014 1). xy)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(viii) Given expression -26, \u2013 24. -22, to 36 terms<\/p>\n\n\n\n

First term (a) = -26<\/p>\n\n\n\n

Common difference (d) = -24 \u2013 (-26)<\/p>\n\n\n\n

= -24 + 26 = 2<\/p>\n\n\n\n

Sum of n terms, Sn<\/sub> = n\/2{2a + (n \u2013 1)d) for n = 36<\/p>\n\n\n\n

Sn<\/sub> = 36\/2{2(-26) + (36 \u2013 1)2}<\/p>\n\n\n\n

= 18[-52 + 70]<\/p>\n\n\n\n

= 18\u00d718<\/p>\n\n\n\n

= 324<\/p>\n\n\n\n

\u2234 Sn<\/sub> = 324<\/p>\n\n\n\n

2. Find the sum to n terms of the A.P. 5, 2, \u20131, \u2013 4, \u20137, \u2026<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given AP is 5, 2, -1, -4, -7, \u2026..<\/p>\n\n\n\n

Here, a = 5, d = 2 \u2013 5 = -3<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Sn<\/sub> = n\/2{2a + (n \u2013 1)d}<\/p>\n\n\n\n

= n\/2{2.5 + (n \u2013 1) \u2013 3}<\/p>\n\n\n\n

= n\/2{10 \u2013 3(n \u2013 1)}<\/p>\n\n\n\n

= n\/2{13 \u2013 3n)<\/p>\n\n\n\n

\u2234 Sn<\/sub> = n\/2(13 \u2013 3n)<\/p>\n\n\n\n

3. Find the sum of n terms of an A.P. whose the terms is given by an<\/sub> = 5 \u2013 6n.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given nth term of the A.P as an<\/sub> = 5 \u2013 6n<\/p>\n\n\n\n

Put n = 1, we get<\/p>\n\n\n\n

a1<\/sub> = 5 \u2013 6.1 = -1<\/p>\n\n\n\n

So, first term (a) = -1<\/p>\n\n\n\n

Last term (an<\/sub>) = 5 \u2013 6n = 1<\/p>\n\n\n\n

Then, Sn<\/sub> = n\/2(-1 + 5 \u2013 6n)<\/p>\n\n\n\n

= n\/2(4 \u2013 6n) = n(2 \u2013 3n)<\/p>\n\n\n\n

4. Find the sum of last ten terms of the A.P. : 8, 10, 12, 14, .. , 126<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given A.P. 8, 10, 12, 14, .. , 126<\/p>\n\n\n\n

Here, a = 8 , d = 10 \u2013 8 = 2<\/p>\n\n\n\n

We know that, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So, to find the number of terms<\/p>\n\n\n\n

126 = 8 + (n \u2013 1)2<\/p>\n\n\n\n

126 = 8 + 2n \u2013 2<\/p>\n\n\n\n

2n = 120<\/p>\n\n\n\n

n = 60<\/p>\n\n\n\n

Next, let\u2019s find the 51st<\/sup> term<\/p>\n\n\n\n

a51<\/sub> = 8 + 50(2) = 108<\/p>\n\n\n\n

So, the sum of last ten terms is the sum of a51<\/sub> + a52<\/sub> + a53<\/sub> + \u2026\u2026. + a60<\/sub><\/p>\n\n\n\n

Here, n = 10, a = 108 and l = 126<\/p>\n\n\n\n

S = 10\/2 [108 + 126]<\/p>\n\n\n\n

= 5(234)<\/p>\n\n\n\n

= 1170<\/p>\n\n\n\n

Hence, the sum of last ten terms of the A.P is 1170.<\/p>\n\n\n\n

5. Find the sum of first 15 terms of each of the following sequences having nth<\/sup> term as:<\/strong><\/p>\n\n\n\n

(i) an<\/sub> = 3 + 4n <\/strong><\/p>\n\n\n\n

(ii) bn<\/sub> = 5 + 2n <\/strong><\/p>\n\n\n\n

(iii) xn<\/sub> = 6 \u2013 n <\/strong><\/p>\n\n\n\n

(iv) yn<\/sub> = 9 \u2013 5n<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given an A.P. whose nth<\/sup> term is given by an<\/sub> = 3 + 4n<\/p>\n\n\n\n

To find the sum of the n terms of the given A.P., using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n(a + l)\/ 2<\/p>\n\n\n\n

Where, a = the first term l = the last term.<\/p>\n\n\n\n

Putting n = 1 in the given an<\/sub>, we get<\/p>\n\n\n\n

a = 3 + 4(1) = 3 + 4 = 7<\/p>\n\n\n\n

For the last term (l), here n = 15<\/p>\n\n\n\n

a15<\/sub> = 3 + 4(15) = 63<\/p>\n\n\n\n

So, Sn<\/sub> = 15(7 + 63)\/2<\/p>\n\n\n\n

= 15 x 35<\/p>\n\n\n\n

= 525<\/p>\n\n\n\n

Therefore, the sum of the 15 terms of the given A.P. is S15<\/sub> = 525<\/p>\n\n\n\n

(ii) Given an A.P. whose nth<\/sup> term is given by bn<\/sub> = 5 + 2n<\/p>\n\n\n\n

To find the sum of the n terms of the given A.P., using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n(a + l)\/ 2<\/p>\n\n\n\n

Where, a = the first term l = the last term.<\/p>\n\n\n\n

Putting n = 1 in the given bn<\/sub>, we get<\/p>\n\n\n\n

a = 5 + 2(1) = 5 + 2 = 7<\/p>\n\n\n\n

For the last term (l), here n = 15<\/p>\n\n\n\n

a15<\/sub> = 5 + 2(15) = 35<\/p>\n\n\n\n

So, Sn<\/sub> = 15(7 + 35)\/2<\/p>\n\n\n\n

= 15 x 21<\/p>\n\n\n\n

= 315<\/p>\n\n\n\n

Therefore, the sum of the 15 terms of the given A.P. is S15<\/sub> = 315<\/p>\n\n\n\n

(iii) Given an A.P. whose nth<\/sup> term is given by xn<\/sub> = 6 \u2013 n<\/p>\n\n\n\n

To find the sum of the n terms of the given A.P., using the formula<\/p>\n\n\n\n

Sn<\/sub> = n(a + l)\/ 2<\/p>\n\n\n\n

Where, a = the first term l = the last term.<\/p>\n\n\n\n

Putting n = 1 in the given xn<\/sub>, we get<\/p>\n\n\n\n

a = 6 \u2013 1 = 5<\/p>\n\n\n\n

For the last term (l), here n = 15<\/p>\n\n\n\n

a15<\/sub> = 6 \u2013 15 = -9<\/p>\n\n\n\n

So, Sn<\/sub> = 15(5 \u2013 9)\/2<\/p>\n\n\n\n

= 15 x (-2)<\/p>\n\n\n\n

= -30<\/p>\n\n\n\n

Therefore, the sum of the 15 terms of the given A.P. is S15<\/sub> = -30<\/p>\n\n\n\n

(iv) Given an A.P. whose nth<\/sup> term is given by yn<\/sub> = 9 \u2013 5n<\/p>\n\n\n\n

To find the sum of the n terms of the given A.P., using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n(a + l)\/ 2<\/p>\n\n\n\n

Where, a = the first term l = the last term.<\/p>\n\n\n\n

Putting n = 1 in the given yn<\/sub>, we get<\/p>\n\n\n\n

a = 9 \u2013 5(1) = 9 \u2013 5 = 4<\/p>\n\n\n\n

For the last term (l), here n = 15<\/p>\n\n\n\n

a15<\/sub> = 9 \u2013 5(15) = -66<\/p>\n\n\n\n

So, Sn<\/sub> = 15(4 \u2013 66)\/2<\/p>\n\n\n\n

= 15 x (-31)<\/p>\n\n\n\n

= -465<\/p>\n\n\n\n

Therefore, the sum of the 15 terms of the given A.P. is S15<\/sub> = -465<\/p>\n\n\n\n

6. Find the sum of first 20 terms the sequence whose nth<\/sup> term is an<\/sub> = An + B.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given an A.P. whose nth term is given by, an<\/sub> = An + B<\/p>\n\n\n\n

We need to find the sum of first 20 terms.<\/p>\n\n\n\n

To find the sum of the n terms of the given A.P., we use the formula,<\/p>\n\n\n\n

Sn<\/sub> = n(a + l)\/ 2<\/p>\n\n\n\n

Where, a = the first term l = the last term,<\/p>\n\n\n\n

Putting n = 1 in the given an<\/sub>, we get<\/p>\n\n\n\n

a = A(1) + B = A + B<\/p>\n\n\n\n

For the last term (l), here n = 20<\/p>\n\n\n\n

A20<\/sub> = A(20) + B = 20A + B<\/p>\n\n\n\n

S20<\/sub> = 20\/2((A + B) + 20A + B)<\/p>\n\n\n\n

= 10[21A + 2B]<\/p>\n\n\n\n

= 210A + 20B<\/p>\n\n\n\n

Therefore, the sum of the first 20 terms of the given A.P. is 210 A + 20B<\/p>\n\n\n\n

7. Find the sum of first 25 terms of an A.P whose nth<\/sup> term is given by an<\/sub> = 2 \u2013 3n.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given an A.P. whose nth<\/sup> term is given by an<\/sub> = 2 \u2013 3n<\/p>\n\n\n\n

To find the sum of the n terms of the given A.P., we use the formula,<\/p>\n\n\n\n

Sn<\/sub> = n(a + l)\/ 2<\/p>\n\n\n\n

Where, a = the first term l = the last term.<\/p>\n\n\n\n

Putting n = 1 in the given an<\/sub>, we get<\/p>\n\n\n\n

a = 2 \u2013 3(1) = -1<\/p>\n\n\n\n

For the last term (l), here n = 25<\/p>\n\n\n\n

a25<\/sub> = 2 \u2013 3(25) = -73<\/p>\n\n\n\n

So, Sn<\/sub> = 25(-1 \u2013 73)\/2<\/p>\n\n\n\n

= 25 x (-37)<\/p>\n\n\n\n

= -925<\/p>\n\n\n\n

Therefore, the sum of the 25 terms of the given A.P. is S25<\/sub> = -925<\/p>\n\n\n\n

8. Find the sum of first 25 terms of an A.P whose nth<\/sup> term is given by an<\/sub> = 7 \u2013 3n. <\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given an A.P. whose nth<\/sup> term is given by an<\/sub> = 7 \u2013 3n<\/p>\n\n\n\n

To find the sum of the n terms of the given A.P., we use the formula,<\/p>\n\n\n\n

Sn<\/sub> = n(a + l)\/ 2<\/p>\n\n\n\n

Where, a = the first term l = the last term.<\/p>\n\n\n\n

Putting n = 1 in the given an<\/sub>, we get<\/p>\n\n\n\n

a = 7 \u2013 3(1) = 7 \u2013 3 = 4<\/p>\n\n\n\n

For the last term (l), here n = 25<\/p>\n\n\n\n

a15<\/sub> = 7 \u2013 3(25) = -68<\/p>\n\n\n\n

So, Sn<\/sub> = 25(4 \u2013 68)\/2<\/p>\n\n\n\n

= 25 x (-32)<\/p>\n\n\n\n

= -800<\/p>\n\n\n\n

Therefore, the sum of the 15 terms of the given A.P. is S25<\/sub> = -800<\/p>\n\n\n\n

9. If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given the sum of the certain number of terms of an A.P. = 116<\/p>\n\n\n\n

We know that, Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Where; a = first term for the given A.P.<\/p>\n\n\n\n

d = common difference of the given A.P.<\/p>\n\n\n\n

n = number of terms So for the given A.P.(25, 22, 19,\u2026)<\/p>\n\n\n\n

Here we have, the first term (a) = 25<\/p>\n\n\n\n

The sum of n terms Sn<\/sub> = 116<\/p>\n\n\n\n

Common difference of the A.P. (d) = a2<\/sub> \u2013 a1<\/sub> = 22 \u2013 25 = -3<\/p>\n\n\n\n

Now, substituting values in Sn<\/sub><\/p>\n\n\n\n

\u27f9 116 = n\/2[2(25) + (n \u2212 1)(\u22123)]<\/p>\n\n\n\n

\u27f9 (n\/2)[50 + (\u22123n + 3)]  = 116<\/p>\n\n\n\n

\u27f9 (n\/2)[53 \u2212 3n] = 116<\/p>\n\n\n\n

\u27f9 53n \u2013 3n2<\/sup> = 116 x 2<\/p>\n\n\n\n

Thus, we get the following quadratic equation,<\/p>\n\n\n\n

3n2<\/sup> \u2013 53n + 232 = 0<\/p>\n\n\n\n

By factorization method of solving, we have<\/p>\n\n\n\n

\u27f9 3n2<\/sup> \u2013 24n \u2013 29n + 232 = 0<\/p>\n\n\n\n

\u27f9 3n( n \u2013 8 ) \u2013 29 ( n \u2013 8 ) = 0<\/p>\n\n\n\n

\u27f9 (3n \u2013 29)( n \u2013 8 ) = 0<\/p>\n\n\n\n

So, 3n \u2013 29 = 0<\/p>\n\n\n\n

\u27f9 n = 29\/3<\/p>\n\n\n\n

Also, n \u2013 8 = 0<\/p>\n\n\n\n

\u27f9 n = 8<\/p>\n\n\n\n

Since, n cannot be a fraction, so the number of terms is taken as 8.<\/p>\n\n\n\n

So, the term is:<\/p>\n\n\n\n

a8<\/sub> = a1<\/sub> + 7d = 25 + 7(-3) = 25 \u2013 21 = 4<\/p>\n\n\n\n

Hence, the last term of the given A.P. such that the sum of the terms is 116 is 4.<\/p>\n\n\n\n

10. (i) How many terms of the sequence 18, 16, 14\u2026.  should be taken so that their sum is zero. <\/strong><\/p>\n\n\n\n

(ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40? <\/strong><\/p>\n\n\n\n

(iii) How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636? <\/strong><\/p>\n\n\n\n

(iv) How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693? <\/strong><\/p>\n\n\n\n

(v) How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero? <\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given AP. is 18, 16, 14, \u2026<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Here,<\/p>\n\n\n\n

The first term (a) = 18<\/p>\n\n\n\n

The sum of n terms (Sn<\/sub>) = 0 (given)<\/p>\n\n\n\n

Common difference of the A.P.<\/p>\n\n\n\n

(d) = a2<\/sub>  \u2013 a1<\/sub> = 16 \u2013 18 = \u2013 2<\/p>\n\n\n\n

So, on substituting the values in Sn<\/sub><\/p>\n\n\n\n

\u27f9 0 = n\/2[2(18) + (n \u2212 1)(\u22122)]<\/p>\n\n\n\n

\u27f9 0 = n\/2[36 + (\u22122n + 2)]<\/p>\n\n\n\n

\u27f9 0 = n\/2[38 \u2212 2n] Further, n\/2<\/p>\n\n\n\n

\u27f9 n = 0 Or, 38 \u2013 2n = 0<\/p>\n\n\n\n

\u27f9 2n = 38<\/p>\n\n\n\n

\u27f9 n = 19<\/p>\n\n\n\n

Since, the number of terms cannot be zer0, hence the number of terms (n) should be 19.<\/p>\n\n\n\n

(ii) Given, the first term (a) = -14, Filth term (a5<\/sub>) = 2, Sum of terms (Sn<\/sub>) = 40 of the A.P.<\/p>\n\n\n\n

If the common difference is taken as d.<\/p>\n\n\n\n

Then, a5<\/sub> = a <\/sub>+ 4d<\/p>\n\n\n\n

\u27f9 2 = -14 + 4d<\/p>\n\n\n\n

\u27f9 2 + 14 = 4d<\/p>\n\n\n\n

\u27f9 4d = 16<\/p>\n\n\n\n

\u27f9 d = 4<\/p>\n\n\n\n

Next, we know that Sn <\/sub>= n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Where; a = first term for the given A.P.<\/p>\n\n\n\n

d = common difference of the given A.P.<\/p>\n\n\n\n

n = number of terms<\/p>\n\n\n\n

Now, on substituting the values in Sn<\/sub><\/p>\n\n\n\n

\u27f9 40 = n\/2[2(\u221214) + (n \u2212 1)(4)]<\/p>\n\n\n\n

\u27f9 40 = n\/2[\u221228 + (4n \u2212 4)]<\/p>\n\n\n\n

\u27f9 40 = n\/2[\u221232 + 4n]<\/p>\n\n\n\n

\u27f9 40(2) = \u2013 32n + 4n2<\/sup><\/p>\n\n\n\n

So, we get the following quadratic equation,<\/p>\n\n\n\n

4n2<\/sup> \u2013 32n \u2013 80 = 0<\/p>\n\n\n\n

\u27f9 n2<\/sup> \u2013 8n \u2013 20 = 0<\/p>\n\n\n\n

On solving by factorization method, we get<\/p>\n\n\n\n

n2<\/sup> \u2013 10n + 2n \u2013 20 = 0<\/p>\n\n\n\n

\u27f9 n(n \u2013 10) + 2( n \u2013 10 ) = 0<\/p>\n\n\n\n

\u27f9 (n + 2)(n \u2013 10) = 0<\/p>\n\n\n\n

Either, n + 2 = 0<\/p>\n\n\n\n

\u27f9 n = -2<\/p>\n\n\n\n

Or, n \u2013 10 = 0<\/p>\n\n\n\n

\u27f9 n = 10<\/p>\n\n\n\n

Since the number of terms cannot be negative.<\/strong><\/p>\n\n\n\n

Therefore, the number of terms (n) is 10.<\/p>\n\n\n\n

(iii) Given AP is 9, 17, 25,\u2026<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Here we have,<\/p>\n\n\n\n

The first term (a) = 9 and the sum of n terms (Sn<\/sub>) = 636<\/p>\n\n\n\n

Common difference of the A.P. (d) = a2<\/sub>  \u2013 a1<\/sub> = 17 \u2013 9 = 8<\/p>\n\n\n\n

Substituting the values in Sn<\/sub>, we get<\/p>\n\n\n\n

\u27f9 636 = n\/2[2(9) + (n \u2212 1)(8)]<\/p>\n\n\n\n

\u27f9 636 = n\/2[18 + (8n \u2212 8)]<\/p>\n\n\n\n

\u27f9 636(2) = (n)[10 + 8n]<\/p>\n\n\n\n

\u27f9 1271 = 10n + 8n2<\/sup><\/p>\n\n\n\n

Now, we get the following quadratic equation,<\/p>\n\n\n\n

\u27f9 8n2<\/sup> + 10n \u2013 1272 = 0<\/p>\n\n\n\n

\u27f9 4n2<\/sup>+ 5n \u2013 636 = 0<\/p>\n\n\n\n

On solving by factorisation method, we have<\/p>\n\n\n\n

\u27f9 4n2<\/sup> \u2013 48n + 53n \u2013 636 = 0<\/p>\n\n\n\n

\u27f9 4n(n \u2013 12) + 53(n \u2013 12) = 0<\/p>\n\n\n\n

\u27f9 (4n + 53)(n \u2013 12) = 0<\/p>\n\n\n\n

Either 4n + 53 = 0 \u27f9 n = -53\/4<\/p>\n\n\n\n

Or, n \u2013 12 = 0 \u27f9 n = 12<\/p>\n\n\n\n

Since, the number of terms cannot be a fraction.<\/p>\n\n\n\n

Therefore, the number of terms (n) is 12.<\/p>\n\n\n\n

(iv) Given A.P. is 63, 60, 57,\u2026<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Here we have,<\/p>\n\n\n\n

the first term (a) = 63<\/p>\n\n\n\n

The sum of n terms (Sn<\/sub>) = 693<\/p>\n\n\n\n

Common difference of the A.P. (d) = a2<\/sub> \u2013 a1<\/sub> = 60 \u2013 63 = \u20133<\/p>\n\n\n\n

On substituting the values in Sn <\/sub>we get<\/p>\n\n\n\n

\u27f9 693 = n\/2[2(63) + (n \u2212 1)(\u22123)]<\/p>\n\n\n\n

\u27f9 693 = n\/2[126+(\u22123n + 3)]<\/p>\n\n\n\n

\u27f9 693 = n\/2[129 \u2212 3n]<\/p>\n\n\n\n

\u27f9 693(2) = 129n \u2013 3n2<\/sup><\/p>\n\n\n\n

Now, we get the following quadratic equation.<\/p>\n\n\n\n

\u27f9 3n2<\/sup> \u2013 129n + 1386 = 0<\/p>\n\n\n\n

\u27f9 n2<\/sup> \u2013 43n + 462<\/p>\n\n\n\n

Solving by factorisation method, we have<\/p>\n\n\n\n

\u27f9 n2<\/sup> \u2013 22n \u2013 21n + 462 = 0<\/p>\n\n\n\n

\u27f9 n(n \u2013 22) -21(n \u2013 22) = 0<\/p>\n\n\n\n

\u27f9 (n \u2013 22) (n \u2013 21) = 0<\/p>\n\n\n\n

Either, n \u2013 22 = 0 \u27f9 n = 22<\/p>\n\n\n\n

Or,  n \u2013 21 = 0 \u27f9 n = 21<\/p>\n\n\n\n

Now, the 22nd<\/sup> term will be a22<\/sub> = a1<\/sub> + 21d = 63 + 21( -3 ) = 63 \u2013 63 = 0<\/p>\n\n\n\n

So, the sum of 22 as well as 21 terms is 693.<\/p>\n\n\n\n

Therefore, the number of terms (n) is 21 or 22.<\/p>\n\n\n\n

(v) Given A.P. is 27, 24, 21. . .<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Here we have, the first term (a) = 27<\/p>\n\n\n\n

The sum of n terms (Sn<\/sub>) = 0<\/p>\n\n\n\n

Common difference of the A.P. (d) = a2<\/sub> \u2013 a1<\/sub> = 24 \u2013 27 = -3<\/p>\n\n\n\n

On substituting the values in Sn<\/sub>, we get<\/p>\n\n\n\n

\u27f9 0 = n\/2[2(27) + (n \u2212 1)( \u2212 3)]<\/p>\n\n\n\n

\u27f9 0 = (n)[54 + (n \u2013 1)(-3)]<\/p>\n\n\n\n

\u27f9 0 = (n)[54 \u2013 3n + 3]<\/p>\n\n\n\n

\u27f9 0 = n [57 \u2013 3n] Further we have, n = 0 Or, 57 \u2013 3n = 0<\/p>\n\n\n\n

\u27f9 3n = 57<\/p>\n\n\n\n

\u27f9 n = 19<\/p>\n\n\n\n

The number of terms cannot be zero,<\/p>\n\n\n\n

Hence, the numbers of terms (n) is 19.<\/p>\n\n\n\n

11. Find the sum of the first<\/strong><\/p>\n\n\n\n

(i) 11 terms of the A.P. : 2, 6, 10, 14,  . . . <\/strong><\/p>\n\n\n\n

(ii) 13 terms of the A.P. : -6, 0, 6, 12, . . . <\/strong><\/p>\n\n\n\n

(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the sum of terms for different arithmetic progressions is given by<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms<\/p>\n\n\n\n

(i) Given A.P 2, 6, 10, 14,\u2026 to 11 terms.<\/p>\n\n\n\n

Common difference (d) = a2<\/sub> \u2013 a1<\/sub> = 10 \u2013 6 = 4<\/p>\n\n\n\n

Number of terms (n) = 11<\/p>\n\n\n\n

First term for the given A.P. (a) = 2<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

S11<\/sub>  = 11\/2[2(2) + (11 \u2212 1)4]<\/p>\n\n\n\n

= 11\/2[2(2) + (10)4]<\/p>\n\n\n\n

= 11\/2[4 + 40]<\/p>\n\n\n\n

= 11 \u00d7 22<\/p>\n\n\n\n

= 242<\/p>\n\n\n\n

Hence, the sum of first 11 terms for the given A.P. is 242<\/p>\n\n\n\n

(ii) Given A.P. \u2013 6, 0, 6, 12, \u2026 to 13 terms.<\/p>\n\n\n\n

Common difference (d) = a2<\/sub> \u2013 a1<\/sub> = 6 \u2013 0 = 6<\/p>\n\n\n\n

Number of terms (n) = 13<\/p>\n\n\n\n

First term (a) = -6<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

S13<\/sub>  = 13\/2[2(\u2212 6) + (13 \u20131)6]<\/p>\n\n\n\n

= 13\/2[(\u221212) + (12)6]<\/p>\n\n\n\n

= 13\/2[60] = 390<\/p>\n\n\n\n

Hence, the sum of first 13 terms for the given A.P. is 390<\/p>\n\n\n\n

(iii) 51 terms of an AP whose a2<\/sub> = 2 and a4<\/sub> = 8<\/p>\n\n\n\n

We know that, a2<\/sub> = a + d<\/p>\n\n\n\n

2 = a + d  \u2026(2)<\/p>\n\n\n\n

Also, a4<\/sub> = a + 3d<\/p>\n\n\n\n

8 = a + 3d  \u2026 (2)<\/p>\n\n\n\n

Subtracting (1) from (2), we have<\/p>\n\n\n\n

2d = 6<\/p>\n\n\n\n

d = 3<\/p>\n\n\n\n

Substituting d = 3 in (1), we get<\/p>\n\n\n\n

2 = a + 3<\/p>\n\n\n\n

\u27f9 a = -1<\/p>\n\n\n\n

Given that the number of terms (n) = 51<\/p>\n\n\n\n

First term (a) = -1<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Sn<\/sub>  = 51\/2[2(\u22121) + (51 \u2212 1)(3)]<\/p>\n\n\n\n

= 51\/2[\u22122 + 150]<\/p>\n\n\n\n

= 51\/2[148]<\/p>\n\n\n\n

= 3774<\/p>\n\n\n\n

Hence, the sum of first 51 terms for the A.P. is 3774.<\/p>\n\n\n\n

12. Find the sum of<\/strong><\/p>\n\n\n\n

(i) the first 15 multiples of 8<\/strong><\/p>\n\n\n\n

(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.<\/strong><\/p>\n\n\n\n

(iii) all 3 \u2013 digit natural numbers which are divisible by 13.<\/strong><\/p>\n\n\n\n

(iv) all 3 \u2013 digit natural numbers which are multiples of 11.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the sum of terms for an A.P is given by<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms<\/p>\n\n\n\n

(i) Given, first 15 multiples of 8.<\/p>\n\n\n\n

These multiples form an A.P: 8, 16, 24, \u2026\u2026 , 120<\/p>\n\n\n\n

Here, a = 8 , d = 61 \u2013 8 = 8 and the number of terms(n) = 15<\/p>\n\n\n\n

Now, finding the sum of 15 terms, we have<\/p>\n\n\n\n

\\<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence, the sum of the first 15 multiples of 8 is 960<\/p>\n\n\n\n

(ii)(a) First 40 positive integers divisible by 3.<\/p>\n\n\n\n

Hence, the first multiple is 3 and the 40th<\/sup> multiple is 120.<\/p>\n\n\n\n

And, these terms will form an A.P. with the common difference of 3.<\/p>\n\n\n\n

Here, First term (a) = 3<\/p>\n\n\n\n

Number of terms (n) = 40<\/p>\n\n\n\n

Common difference (d) = 3<\/p>\n\n\n\n

So, the sum of 40 terms<\/p>\n\n\n\n

S40<\/sub>  = 40\/2[2(3) + (40 \u2212 1)3]<\/p>\n\n\n\n

= 20[6 + (39)3]<\/p>\n\n\n\n

= 20(6 + 117)<\/p>\n\n\n\n

= 20(123) = 2460<\/p>\n\n\n\n

Thus, the sum of first 40 multiples of 3 is 2460.<\/p>\n\n\n\n

(b) First 40 positive integers divisible by 5<\/p>\n\n\n\n

Hence, the first multiple is 5 and the 40th<\/sup> multiple is 200.<\/p>\n\n\n\n

And, these terms will form an A.P. with the common difference of 5.<\/p>\n\n\n\n

Here, First term (a) = 5<\/p>\n\n\n\n

Number of terms (n) = 40<\/p>\n\n\n\n

Common difference (d) = 5<\/p>\n\n\n\n

So, the sum of 40 terms<\/p>\n\n\n\n

S40  <\/sub>= 40\/2[2(5) + (40 \u2212 1)5]<\/p>\n\n\n\n

= 20[10 + (39)5]<\/p>\n\n\n\n

= 20 (10 + 195)<\/p>\n\n\n\n

= 20 (205) = 4100<\/p>\n\n\n\n

Hence, the sum of first 40 multiples of 5 is 4100.<\/p>\n\n\n\n

(c) First 40 positive integers divisible by 6<\/p>\n\n\n\n

Hence, the first multiple is 6 and the 40th<\/sup> multiple is 240.<\/p>\n\n\n\n

And, these terms will form an A.P. with the common difference of 6.<\/p>\n\n\n\n

Here, First term (a) = 6<\/p>\n\n\n\n

Number of terms (n) = 40<\/p>\n\n\n\n

Common difference (d) = 6<\/p>\n\n\n\n

So, the sum of 40 terms<\/p>\n\n\n\n

S40<\/sub>  = 40\/2[2(6) + (40 \u2212 1)6]<\/p>\n\n\n\n

= 20[12 + (39)6]<\/p>\n\n\n\n

=20(12 + 234)<\/p>\n\n\n\n

= 20(246) = 4920<\/p>\n\n\n\n

Hence, the sum of first 40 multiples of 6 is 4920.<\/p>\n\n\n\n

(iii) All 3 digit natural number which are divisible by 13.<\/p>\n\n\n\n

So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.<\/p>\n\n\n\n

And, these terms form an A.P. with the common difference of 13.<\/p>\n\n\n\n

Here, first term (a) = 104 and the last term (l) = 988<\/p>\n\n\n\n

Common difference (d) = 13<\/p>\n\n\n\n

Finding the number of terms in the A.P. by, an<\/sub> = a + (n \u2212 1)d<\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

988 = 104 + (n \u2013 1)13<\/p>\n\n\n\n

\u27f9 988 = 104 + 13n -13<\/p>\n\n\n\n

\u27f9 988 = 91 + 13n<\/p>\n\n\n\n

\u27f9 13n = 897<\/p>\n\n\n\n

\u27f9 n = 69<\/p>\n\n\n\n

Now, using the formula for the sum of n terms, we get<\/p>\n\n\n\n

S69<\/sub>  = 69\/2[2(104) + (69 \u2212 1)13]<\/p>\n\n\n\n

= 69\/2[208 + 884]<\/p>\n\n\n\n

= 69\/2[1092]<\/p>\n\n\n\n

= 69(546)<\/p>\n\n\n\n

= 37674<\/p>\n\n\n\n

Hence, the sum of all 3 digit multiples of 13 is 37674.<\/p>\n\n\n\n

(iv) All 3 digit natural number which are multiples of 11.<\/p>\n\n\n\n

So, we know that the first 3 digit multiple of 11 is 110 and the last 3 digit multiple of 13 is 990.<\/p>\n\n\n\n

And, these terms form an A.P. with the common difference of 11.<\/p>\n\n\n\n

Here, first term (a) = 110 and the last term (l) = 990<\/p>\n\n\n\n

Common difference (d) = 11<\/p>\n\n\n\n

Finding the number of terms in the A.P. by, an<\/sub> = a + (n \u2212 1)d<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

990 = 110 + (n \u2013 1)11<\/p>\n\n\n\n

\u27f9 990 = 110 + 11n -11<\/p>\n\n\n\n

\u27f9 990 = 99 + 11n<\/p>\n\n\n\n

\u27f9 11n = 891<\/p>\n\n\n\n

\u27f9 n = 81<\/p>\n\n\n\n

Now, using the formula for the sum of n terms, we get<\/p>\n\n\n\n

S81<\/sub>  = 81\/2[2(110) + (81 \u2212 1)11]<\/p>\n\n\n\n

= 81\/2[220 + 880]<\/p>\n\n\n\n

= 81\/2[1100]<\/p>\n\n\n\n

= 81(550)<\/p>\n\n\n\n

= 44550<\/p>\n\n\n\n

Hence, the sum of all 3 digit multiples of 11 is 44550.<\/p>\n\n\n\n

13. Find the sum:<\/strong><\/p>\n\n\n\n

(i) 2 + 4 + 6 + . . . + 200 <\/strong><\/p>\n\n\n\n

(ii) 3 + 11 + 19 + . . . + 803 <\/strong><\/p>\n\n\n\n

(iii) (-5) + (-8) + (-11) + . . . + (- 230) <\/strong><\/p>\n\n\n\n

(iv) 1 + 3 + 5 + 7 + . . . + 199 <\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(vi) 34 + 32 + 30 + . . . + 10 <\/strong><\/p>\n\n\n\n

(vii) 25 + 28 + 31 + . . . + 100 <\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the sum of terms for an A.P is given by<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms<\/p>\n\n\n\n

Or Sn<\/sub> = n\/2[a + l]<\/p>\n\n\n\n

Where; a = first term for the given A.P. ;l = last term for the given A.P<\/p>\n\n\n\n

(i) Given series. 2 + 4 + 6 + . . . + 200 which is an A.P<\/p>\n\n\n\n

Where, a = 2 ,d = 4 \u2013 2 = 2 and last term (an <\/sub>= l) = 200<\/p>\n\n\n\n

We know that, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

200 = 2 + (n \u2013 1)2<\/p>\n\n\n\n

200 = 2 + 2n \u2013 2<\/p>\n\n\n\n

n = 200\/2 = 100<\/p>\n\n\n\n

Now, for the sum of these 100 terms<\/p>\n\n\n\n

S100 <\/sub>= 100\/2 [2 + 200]<\/p>\n\n\n\n

= 50(202)<\/p>\n\n\n\n

= 10100<\/p>\n\n\n\n

Hence, the sum of terms of the given series is 10100.<\/p>\n\n\n\n

(ii) Given series. 3 + 11 + 19 + . . . + 803 which is an A.P<\/p>\n\n\n\n

Where, a = 3 ,d = 11 \u2013 3 = 8 and last term (an <\/sub>= l) = 803<\/p>\n\n\n\n

We know that, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

803 = 3 + (n \u2013 1)8<\/p>\n\n\n\n

803 = 3 + 8n \u2013 8<\/p>\n\n\n\n

n = 808\/8 = 101<\/p>\n\n\n\n

Now, for the sum of these 101 terms<\/p>\n\n\n\n

S101 <\/sub>= 101\/2 [3 + 803]<\/p>\n\n\n\n

= 101(806)\/2<\/p>\n\n\n\n

= 101 x 403<\/p>\n\n\n\n

= 40703<\/p>\n\n\n\n

Hence, the sum of terms of the given series is 40703.<\/p>\n\n\n\n

(iii) Given series (-5) + (-8) + (-11) + . . . + (- 230) <\/strong>which is an A.P<\/p>\n\n\n\n

Where, a = -5 ,d = -8 \u2013 (-5) = -3 and last term (an <\/sub>= l) = -230<\/p>\n\n\n\n

We know that, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

-230 = -5 + (n \u2013 1)(-3)<\/p>\n\n\n\n

-230 = -5 \u2013 3n + 3<\/p>\n\n\n\n

3n = -2 + 230<\/p>\n\n\n\n

n = 228\/3 = 76<\/p>\n\n\n\n

Now, for the sum of these 76 terms<\/p>\n\n\n\n

S76 <\/sub>= 76\/2 [-5 + (-230)]<\/p>\n\n\n\n

= 38 x (-235)<\/p>\n\n\n\n

= -8930<\/p>\n\n\n\n

Hence, the sum of terms of the given series is -8930.<\/p>\n\n\n\n

(iv) Given series. 1 + 3 + 5 + 7 + . . . + 199 <\/strong>which is an A.P<\/p>\n\n\n\n

Where, a = 1 ,d = 3 \u2013 1 = 2 and last term (an <\/sub>= l) = 199<\/p>\n\n\n\n

We know that, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

199 = 1 + (n \u2013 1)2<\/p>\n\n\n\n

199 = 1 + 2n \u2013 2<\/p>\n\n\n\n

n = 200\/2 = 100<\/p>\n\n\n\n

Now, for the sum of these 100 terms<\/p>\n\n\n\n

S100 <\/sub>= 100\/2 [1 + 199]<\/p>\n\n\n\n

= 50(200)<\/p>\n\n\n\n

= 10000<\/p>\n\n\n\n

Hence, the sum of terms of the given series is 10000.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(v) Given series which is an A.P<\/p>\n\n\n\n

Where, a = 7, d = 10 \u00bd \u2013 7 = (21 \u2013 14)\/2 = 7\/2 and last term (an <\/sub>= l) = 84<\/p>\n\n\n\n

We know that, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

84 = 7 + (n \u2013 1)(7\/2)<\/p>\n\n\n\n

168 = 14 + 7n \u2013 7<\/p>\n\n\n\n

n = (168 \u2013 7)\/7 = 161\/7 = 23<\/p>\n\n\n\n

Now, for the sum of these 23 terms<\/p>\n\n\n\n

S23 <\/sub>= 23\/2 [7 + 84]<\/p>\n\n\n\n

= 23(91)\/2<\/p>\n\n\n\n

= 2093\/2<\/p>\n\n\n\n

Hence, the sum of terms of the given series is 2093\/2.<\/p>\n\n\n\n

(vi) Given series, 34 + 32 + 30 + . . . + 10 <\/strong>which is an A.P<\/p>\n\n\n\n

Where, a = 34 ,d = 32 \u2013 34 = -2 and last term (an <\/sub>= l) = 10<\/p>\n\n\n\n

We know that, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

10 = 34 + (n \u2013 1)(-2)<\/p>\n\n\n\n

10 = 34 \u2013 2n + 2<\/p>\n\n\n\n

n = (36 \u2013 10)\/2 = 13<\/p>\n\n\n\n

Now, for the sum of these 13 terms<\/p>\n\n\n\n

S13 <\/sub>= 13\/2 [34 + 10]<\/p>\n\n\n\n

= 13(44)\/2<\/p>\n\n\n\n

= 13 x 22<\/p>\n\n\n\n

= 286<\/p>\n\n\n\n

Hence, the sum of terms of the given series is 286.<\/p>\n\n\n\n

(vii) Given series, 25 + 28 + 31 + . . . + 100 <\/strong>which is an A.P<\/p>\n\n\n\n

Where, a = 25 ,d = 28 \u2013 25 = 3 and last term (an <\/sub>= l) = 100<\/p>\n\n\n\n

We know that, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

100 = 25 + (n \u2013 1)(3)<\/p>\n\n\n\n

100 = 25 + 3n \u2013 3<\/p>\n\n\n\n

n = (100 \u2013 22)\/3 = 26<\/p>\n\n\n\n

Now, for the sum of these 26 terms<\/p>\n\n\n\n

S100 <\/sub>= 26\/2 [25 + 100]<\/p>\n\n\n\n

= 13(125)<\/p>\n\n\n\n

= 1625<\/p>\n\n\n\n

Hence, the sum of terms of the given series is 1625.<\/p>\n\n\n\n

14. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, the first term of the A.P (a) = 17<\/p>\n\n\n\n

The last term of the A.P (l) = 350<\/p>\n\n\n\n

The common difference (d) of the A.P. = 9<\/p>\n\n\n\n

Let the number of terms be n. And, we know that; l = a + (n \u2013 1)d<\/p>\n\n\n\n

So, 350 = 17 + (n- 1) 9<\/p>\n\n\n\n

\u27f9 350 = 17 + 9n \u2013 9<\/p>\n\n\n\n

\u27f9 350 = 8 + 9n<\/p>\n\n\n\n

\u27f9 350 \u2013 8 = 9n<\/p>\n\n\n\n

Thus we get, n = 38<\/p>\n\n\n\n

Now, finding the sum of terms<\/p>\n\n\n\n

Sn<\/sub>  = n\/2[a + l]<\/p>\n\n\n\n

= 38\/2(17 + 350)<\/p>\n\n\n\n

= 19 \u00d7 367<\/p>\n\n\n\n

= 6973<\/p>\n\n\n\n

Hence, the number of terms is of the A.P is 38 and their sum is 6973.<\/p>\n\n\n\n

15. The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s consider the first term as a and the common difference as d.<\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a3<\/sub> = 7 \u2026. (1) and,<\/p>\n\n\n\n

a7<\/sub> = 3a3<\/sub> + 2   \u2026. (2)<\/p>\n\n\n\n

So, using (1) in (2), we get,<\/p>\n\n\n\n

a7 <\/sub>= 3(7) + 2 = 21 + 2 = 23  \u2026. (3)<\/p>\n\n\n\n

Also, we know that<\/p>\n\n\n\n

an<\/sub> = a +(n \u2013 1)d<\/p>\n\n\n\n

So, the 3th term (for n = 3),<\/p>\n\n\n\n

a3<\/sub> = a + (3 \u2013 1)d<\/p>\n\n\n\n

\u27f9 7 = a + 2d   (Using 1)<\/p>\n\n\n\n

\u27f9 a = 7 \u2013 2d     \u2026. (4)<\/p>\n\n\n\n

Similarly, for the 7th term (n = 7),<\/p>\n\n\n\n

a7<\/sub> = a + (7 \u2013 1) d 24 = a + 6d = 23  (Using 3)<\/p>\n\n\n\n

a = 23 \u2013 6d  \u2026. (5)<\/p>\n\n\n\n

Subtracting (4) from (5), we get,<\/p>\n\n\n\n

a \u2013 a = (23 \u2013 6d) \u2013 (7 \u2013 2d)<\/p>\n\n\n\n

\u27f9 0 = 23 \u2013 6d \u2013 7 + 2d<\/p>\n\n\n\n

\u27f9 0 = 16 \u2013 4d<\/p>\n\n\n\n

\u27f9 4d = 16<\/p>\n\n\n\n

\u27f9 d = 4<\/p>\n\n\n\n

Now, to find a, we substitute the value of d in  (4), a =7 \u2013 2(4)<\/p>\n\n\n\n

\u27f9 a = 7 \u2013 8<\/p>\n\n\n\n

a = -1<\/p>\n\n\n\n

Hence, for the A.P. a = -1 and d = 4<\/p>\n\n\n\n

For finding the sum, we know that<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d] and n = 20 (given)<\/p>\n\n\n\n

S20<\/sub>  = 20\/2[2(\u22121) + (20 \u2212 1)(4)]<\/p>\n\n\n\n

= (10)[-2 + (19)(4)]<\/p>\n\n\n\n

= (10)[-2 + 76]<\/p>\n\n\n\n

= (10)[74]<\/p>\n\n\n\n

= 740<\/p>\n\n\n\n

Hence, the sum of first 20 terms for the given A.P. is 740<\/p>\n\n\n\n

16. The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

The first term of the A.P (a) = 2<\/p>\n\n\n\n

The last term of the A.P (l) = 50<\/p>\n\n\n\n

Sum of all the terms Sn<\/sub> = 442<\/p>\n\n\n\n

So, let the common difference of the A.P. be taken as d.<\/p>\n\n\n\n

The sum of all the terms is given as,<\/p>\n\n\n\n

442 = (n\/2)(2 + 50)<\/p>\n\n\n\n

\u27f9 442 = (n\/2)(52)<\/p>\n\n\n\n

\u27f9 26n = 442<\/p>\n\n\n\n

\u27f9 n = 17<\/p>\n\n\n\n

Now, the last term is expressed as<\/p>\n\n\n\n

50 = 2 + (17 \u2013 1)d<\/p>\n\n\n\n

\u27f9 50 = 2 + 16d<\/p>\n\n\n\n

\u27f9 16d = 48<\/p>\n\n\n\n

\u27f9 d = 3<\/p>\n\n\n\n

Thus, the common difference of the A.P. is d = 3.<\/p>\n\n\n\n

17. If 12th<\/sup> term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us take the first term as a and the common difference as d.<\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

a12<\/sub> = -13 S4<\/sub> = 24<\/p>\n\n\n\n

Also, we know that an <\/sub>= a + (n \u2013 1)d<\/p>\n\n\n\n

So, for the 12th term<\/p>\n\n\n\n

a12<\/sub> = a + (12 \u2013 1)d = -13<\/p>\n\n\n\n

\u27f9 a + 11d = -13<\/p>\n\n\n\n

a = -13 \u2013 11d  \u2026. (1)<\/p>\n\n\n\n

And, we that for sum of terms<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Here, n = 4<\/p>\n\n\n\n

S4<\/sub>  = 4\/2[2(a) + (4 \u2212 1)d]<\/p>\n\n\n\n

\u27f9 24 = (2)[2a + (3)(d)]<\/p>\n\n\n\n

\u27f9 24 = 4a + 6d<\/p>\n\n\n\n

\u27f9 4a = 24 \u2013 6d<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Subtracting (1) from (2), we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Further simplifying for d, we get,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u27f9 -19 \u00d7 2 = 19d<\/p>\n\n\n\n

\u27f9 d = \u2013 2<\/p>\n\n\n\n

On substituting the value of d in (1), we find a<\/p>\n\n\n\n

a = -13 \u2013 11(-2)<\/p>\n\n\n\n

a = -13 + 22<\/p>\n\n\n\n

a = 9<\/p>\n\n\n\n

Next, the sum of 10 term is given by<\/p>\n\n\n\n

S10<\/sub>  = 10\/2[2(9) + (10 \u2212 1)(\u22122)]<\/p>\n\n\n\n

= (5)[19 + (9)(-2)]<\/p>\n\n\n\n

= (5)(18 \u2013 18) = 0<\/p>\n\n\n\n

Thus, the sum of first 10 terms for the given A.P. is S10<\/sub> = 0.<\/p>\n\n\n\n

18.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

19. In an A.P., if the first term is 22, the common difference is \u2013 4 and the sum to n terms is 64, find n.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that,<\/p>\n\n\n\n

a = 22, d = \u2013 4 and Sn<\/sub> = 64<\/p>\n\n\n\n

Let us consider the number of terms as n.<\/p>\n\n\n\n

For sum of terms in an A.P, we know that<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\u27f9 Sn<\/sub> = n\/2[2(22) + (n \u2212 1)(\u22124)]<\/p>\n\n\n\n

\u27f9 64 = n\/2[2(22) + (n \u2212 1)(\u22124)]<\/p>\n\n\n\n

\u27f9 64(2) = n(48 \u2013 4n)<\/p>\n\n\n\n

\u27f9 128 = 48n \u2013 4n2<\/sup><\/p>\n\n\n\n

After rearranging the terms, we have a quadratic equation<\/p>\n\n\n\n

4n2<\/sup> \u2013 48n + 128 = 0,<\/p>\n\n\n\n

n2<\/sup> \u2013 12n + 32 = 0 [dividing by 4 on both sides]<\/p>\n\n\n\n

n2<\/sup> \u2013 12n + 32 = 0<\/p>\n\n\n\n

Solving by factorisation method,<\/p>\n\n\n\n

n2<\/sup> \u2013 8n \u2013 4n + 32 = 0<\/p>\n\n\n\n

n ( n \u2013 8 ) \u2013 4 ( n \u2013 8 ) = 0<\/p>\n\n\n\n

(n \u2013 8) (n \u2013 4) = 0<\/p>\n\n\n\n

So, we get n \u2013 8 = 0 \u27f9 n = 8<\/p>\n\n\n\n

Or, n \u2013 4 = 0 \u27f9 n = 4<\/p>\n\n\n\n

Hence, the number of terms can be either n = 4 or 8.<\/p>\n\n\n\n

20. In an A.P., if the 5th<\/sup> and 12th<\/sup> terms are 30 and 65 respectively, what is the sum of first 20 terms? <\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s take the first term as a and the common difference to be d<\/p>\n\n\n\n

Given that,<\/p>\n\n\n\n

a5<\/sub> = 30  and a12<\/sub> = 65<\/p>\n\n\n\n

And, we know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a5<\/sub> = a + (5 \u2013 1)d<\/p>\n\n\n\n

30 = a + 4d<\/p>\n\n\n\n

a = 30 \u2013 4d   \u2026. (i)<\/p>\n\n\n\n

Similarly, a12<\/sub> = a + (12 \u2013 1) d<\/p>\n\n\n\n

65 = a + 11d<\/p>\n\n\n\n

a = 65 \u2013 11d \u2026. (ii)<\/p>\n\n\n\n

Subtracting (i) from (ii), we have<\/p>\n\n\n\n

a \u2013 a = (65 \u2013 11d) \u2013 (30 \u2013 4d)<\/p>\n\n\n\n

0 = 65 \u2013 11d \u2013 30 + 4d<\/p>\n\n\n\n

0 = 35 \u2013 7d<\/p>\n\n\n\n

7d = 35<\/p>\n\n\n\n

d = 5<\/p>\n\n\n\n

Putting d in (i), we get<\/p>\n\n\n\n

a = 30 \u2013 4(5)<\/p>\n\n\n\n

a = 30 \u2013 20<\/p>\n\n\n\n

a = 10<\/p>\n\n\n\n

Thus for the A.P; d = 5 and a = 10<\/p>\n\n\n\n

Next, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

Where;<\/p>\n\n\n\n

a = first term of the given A.P.<\/p>\n\n\n\n

d = common difference of the given A.P.<\/p>\n\n\n\n

n = number of terms<\/p>\n\n\n\n

Here n = 20, so we have<\/p>\n\n\n\n

S20 <\/sub>= 20\/2[2(10) + (20 \u2212 1)(5)]<\/p>\n\n\n\n

= (10)[20 + (19)(5)]<\/p>\n\n\n\n

= (10)[20 + 95]<\/p>\n\n\n\n

= (10)[115]<\/p>\n\n\n\n

= 1150<\/p>\n\n\n\n

Hence, the sum of first 20 terms for the given A.P. is 1150<\/p>\n\n\n\n

21. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s take the first term as a and the common difference as d.<\/p>\n\n\n\n

Given that,<\/p>\n\n\n\n

a2<\/sub> = 14 and a3<\/sub> = 18<\/p>\n\n\n\n

And, we know that an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

a2<\/sub> = a + (2 \u2013 1)d<\/p>\n\n\n\n

\u27f9 14 = a + d<\/p>\n\n\n\n

\u27f9 a = 14 \u2013 d \u2026. (i)<\/p>\n\n\n\n

Similarly,<\/p>\n\n\n\n

a3<\/sub> = a + (3 \u2013 1)d<\/p>\n\n\n\n

\u27f9 18 = a + 2d<\/p>\n\n\n\n

\u27f9 a = 18 \u2013 2d \u2026. (ii)<\/p>\n\n\n\n

Subtracting (i) from (ii), we have<\/p>\n\n\n\n

a \u2013 a = (18 \u2013 2d) \u2013 (14 \u2013 d)<\/p>\n\n\n\n

0 = 18 \u2013 2d \u2013 14 + d<\/p>\n\n\n\n

0 = 4 \u2013 d<\/p>\n\n\n\n

d = 4<\/p>\n\n\n\n

Putting d in (i), to find a<\/p>\n\n\n\n

a = 14 \u2013 4<\/p>\n\n\n\n

a = 10<\/p>\n\n\n\n

Thus, for the A.P. d = 4 and a = 10<\/p>\n\n\n\n

Now, to find sum of terms<\/p>\n\n\n\n

Sn<\/sub> = n\/2(2a + (n \u2212 1)d)<\/p>\n\n\n\n

Where,<\/p>\n\n\n\n

a = the first term of the A.P.<\/p>\n\n\n\n

d = common difference of the A.P.<\/p>\n\n\n\n

n = number of terms So, using the formula for<\/p>\n\n\n\n

n = 51,<\/p>\n\n\n\n

\u27f9 S51<\/sub>  = 51\/2[2(10) + (51 \u2013 1)(4)]<\/p>\n\n\n\n

= 51\/2[20 + (40)4]<\/p>\n\n\n\n

= 51\/2[220]<\/p>\n\n\n\n

= 51(110)<\/p>\n\n\n\n

= 5610<\/p>\n\n\n\n

Hence, the sum of the first 51 terms of the given A.P. is 5610<\/p>\n\n\n\n

22. If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

Sum of 7 terms of an A.P. is 49<\/p>\n\n\n\n

\u27f9 S7<\/sub> = 49<\/p>\n\n\n\n

And, sum of 17 terms of an A.P. is 289<\/p>\n\n\n\n

\u27f9 S17<\/sub> = 289<\/p>\n\n\n\n

Let the first term of the A.P be a and common difference as d.<\/p>\n\n\n\n

And, we know that the sum of n terms of an A.P is<\/p>\n\n\n\n

Sn<\/sub> = n\/2[2a + (n \u2212 1)d]<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

S7 <\/sub>= 49 = 7\/2[2a + (7 \u2013 1)d]<\/p>\n\n\n\n

= 7\/2 [2a + 6d]<\/p>\n\n\n\n

= 7[a + 3d]<\/p>\n\n\n\n

\u27f9 7a + 21d = 49<\/p>\n\n\n\n

a + 3d = 7 \u2026.. (i)<\/p>\n\n\n\n

Similarly,<\/p>\n\n\n\n

S17 <\/sub>= 17\/2[2a + (17 \u2013 1)d]<\/p>\n\n\n\n

= 17\/2 [2a + 16d]<\/p>\n\n\n\n

= 17[a + 8d]<\/p>\n\n\n\n

\u27f9 17[a + 8d] = 289<\/p>\n\n\n\n

a + 8d = 17 \u2026.. (ii)<\/p>\n\n\n\n

Now, subtracting (i) from (ii), we have<\/p>\n\n\n\n

a + 8d \u2013 (a + 3d) = 17 \u2013 7<\/p>\n\n\n\n

5d = 10<\/p>\n\n\n\n

d = 2<\/p>\n\n\n\n

Putting d in (i), we find a<\/p>\n\n\n\n

a + 3(2) = 7<\/p>\n\n\n\n

a = 7 \u2013 6 = 1<\/p>\n\n\n\n

So, for the A.P: a = 1 and d = 2<\/p>\n\n\n\n

For the sum of n terms is given by,<\/p>\n\n\n\n

Sn<\/sub>  = n\/2[2(1) + (n \u2212 1)(2)]<\/p>\n\n\n\n

= n\/2[2 + 2n \u2013 2]<\/p>\n\n\n\n

= n\/2[2n]<\/p>\n\n\n\n

= n2<\/sup><\/p>\n\n\n\n

Therefore, the sum of n terms of the A.P is given by n2<\/sup>.<\/p>\n\n\n\n

23. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Sum of first n terms of an A.P is given by Sn<\/sub> = n\/2(2a + (n \u2212 1)d)<\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

First term (a) = 5, last term (an<\/sub>) = 45 and sum of n terms (Sn<\/sub>) = 400<\/p>\n\n\n\n

Now, we know that<\/p>\n\n\n\n

an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u27f9 45 = 5 + (n \u2013 1)d<\/p>\n\n\n\n

\u27f9 40 = nd \u2013 d<\/p>\n\n\n\n

\u27f9 nd \u2013 d = 40 \u2026. (1)<\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

Sn <\/sub>= n\/2(2(a) + (n \u2212 1)d)<\/p>\n\n\n\n

400 = n\/2(2(5) + (n \u2212 1)d)<\/p>\n\n\n\n

800 = n (10 + nd \u2013 d)<\/p>\n\n\n\n

800 = n (10 + 40) [using (1)]<\/p>\n\n\n\n

\u27f9 n = 16<\/p>\n\n\n\n

Putting n in (1), we find d<\/p>\n\n\n\n

nd \u2013 d = 40<\/p>\n\n\n\n

16d \u2013 d = 40<\/p>\n\n\n\n

15d = 40<\/p>\n\n\n\n

d = 8\/3<\/p>\n\n\n\n

Therefore, the common difference of the given A.P. is 8\/3.<\/p>\n\n\n\n

24. In an A.P. the first term is 8, nth<\/sup> term is 33 and the sum of first n term is 123. Find n and the d, the common difference. <\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

The first term of the A.P (a) = 8<\/p>\n\n\n\n

The nth term of the A.P (l) = 33<\/p>\n\n\n\n

And, the sum of all the terms Sn<\/sub> = 123<\/p>\n\n\n\n

Let the common difference of the A.P. be d.<\/p>\n\n\n\n

So, find the number of terms by<\/p>\n\n\n\n

123 = (n\/2)(8 + 33)<\/p>\n\n\n\n

123 = (n\/2)(41)<\/p>\n\n\n\n

n = (123 x 2)\/ 41<\/p>\n\n\n\n

n = 246\/41<\/p>\n\n\n\n

n = 6<\/p>\n\n\n\n

Next, to find the common difference of the A.P. we know that<\/p>\n\n\n\n

l = a + (n \u2013 1)d<\/p>\n\n\n\n

33 = 8 + (6 \u2013 1)d<\/p>\n\n\n\n

33 = 8 + 5d<\/p>\n\n\n\n

5d = 25<\/p>\n\n\n\n

d = 5<\/p>\n\n\n\n

Thus, the number of terms is n = 6 and the common difference of the A.P. is d = 5.<\/p>\n\n\n\n

25. In an A.P. the first term is 22, nth<\/sup> term is -11 and the sum of first n term is 66. Find n and the d, the common difference. <\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

The first term of the A.P (a) = 22<\/p>\n\n\n\n

The nth term of the A.P (l) = -11<\/p>\n\n\n\n

And, sum of all the terms Sn<\/sub> = 66<\/p>\n\n\n\n

Let the common difference of the A.P. be d.<\/p>\n\n\n\n

So, finding the number of terms by<\/p>\n\n\n\n

66 = (n\/2)[22 + (\u221211)]<\/p>\n\n\n\n

66 = (n\/2)[22 \u2212 11]<\/p>\n\n\n\n

(66)(2) = n(11)<\/p>\n\n\n\n

6 \u00d7 2 = n<\/p>\n\n\n\n

n = 12<\/p>\n\n\n\n

Now, for finding d<\/p>\n\n\n\n

We know that, l = a + (n \u2013 1)d<\/p>\n\n\n\n

\u2013 11 = 22 + (12 \u2013 1)d<\/p>\n\n\n\n

-11 = 22 + 11d<\/p>\n\n\n\n

11d = \u2013 33<\/p>\n\n\n\n

d = \u2013 3<\/p>\n\n\n\n

Hence, the number of terms is n = 12 and the common difference d = -3<\/p>\n\n\n\n

26. The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find the common difference.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

First term (a) = 7, last term (an<\/sub>) = 49 and sum of n terms (Sn<\/sub>) = 420<\/p>\n\n\n\n

Now, we know that<\/p>\n\n\n\n

an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u27f9 49 = 7 + (n \u2013 1)d<\/p>\n\n\n\n

\u27f9 43 = nd \u2013 d<\/p>\n\n\n\n

\u27f9 nd \u2013 d = 42  \u2026.. (1)<\/p>\n\n\n\n

Next,<\/p>\n\n\n\n

Sn<\/sub> = n\/2(2(7) + (n \u2212 1)d)<\/p>\n\n\n\n

\u27f9 840 = n[14 + nd \u2013 d]<\/p>\n\n\n\n

\u27f9 840 = n[14 + 42] [using (1)]<\/p>\n\n\n\n

\u27f9 840 = 54n<\/p>\n\n\n\n

\u27f9 n = 15 \u2026. (2)<\/p>\n\n\n\n

So, by substituting (2) in (1), we have<\/p>\n\n\n\n

nd \u2013 d = 42<\/p>\n\n\n\n

\u27f9 15d \u2013 d = 42<\/p>\n\n\n\n

\u27f9 14d = 42<\/p>\n\n\n\n

\u27f9 d = 3<\/p>\n\n\n\n

Therefore, the common difference of the given A.P. is 3.<\/p>\n\n\n\n

27. The first and the last terms of an A.P are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

First term (a) = 5 and the last term (l) = 45<\/p>\n\n\n\n

Also, Sn<\/sub> = 400<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u27f9 45 = 5 + (n \u2013 1)d<\/p>\n\n\n\n

\u27f9 40 = nd \u2013 d<\/p>\n\n\n\n

\u27f9 nd \u2013 d = 40  \u2026.. (1)<\/p>\n\n\n\n

Next,<\/p>\n\n\n\n

Sn<\/sub> = n\/2(2(5) + (n \u2212 1)d)<\/p>\n\n\n\n

\u27f9 400 = n[10 + nd \u2013 d]<\/p>\n\n\n\n

\u27f9 800 = n[10 + 40] [using (1)]<\/p>\n\n\n\n

\u27f9 800 = 50n<\/p>\n\n\n\n

\u27f9 n = 16 \u2026. (2)<\/p>\n\n\n\n

So, by substituting (2) in (1), we have<\/p>\n\n\n\n

nd \u2013 d = 40<\/p>\n\n\n\n

\u27f9 16d \u2013 d = 40<\/p>\n\n\n\n

\u27f9 15d = 40<\/p>\n\n\n\n

\u27f9 d = 8\/3<\/p>\n\n\n\n

Therefore, the common difference of the given A.P. is 8\/3.<\/p>\n\n\n\n

28. The sum of first q terms of an A.P. is 162. The ratio of its 6th<\/sup> term to its 13th<\/sup> term is 1: 2. Find the first and 15th<\/sup> term of the A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let a be the first term and d be the common difference.<\/p>\n\n\n\n

And we know that, sum of first n terms is:<\/p>\n\n\n\n

Sn<\/sub> = n\/2(2a + (n \u2212 1)d)<\/p>\n\n\n\n

Also, nth term is given by an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

From the question, we have<\/p>\n\n\n\n

Sq<\/sub> = 162 and a6 <\/sub>: a13<\/sub> = 1 : 2<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

2a6 <\/sub>= a13<\/sub><\/p>\n\n\n\n

\u27f9 2 [a + (6 \u2013 1d)] = a + (13 \u2013 1)d<\/p>\n\n\n\n

\u27f9 2a + 10d = a + 12d<\/p>\n\n\n\n

\u27f9 a = 2d  \u2026. (1)<\/p>\n\n\n\n

And, S9<\/sub> = 162<\/p>\n\n\n\n

\u27f9 S9<\/sub> = 9\/2(2a + (9 \u2212 1)d)<\/p>\n\n\n\n

\u27f9 162 = 9\/2(2a + 8d)<\/p>\n\n\n\n

\u27f9 162 \u00d7 2 = 9[4d + 8d]  [from (1)]<\/p>\n\n\n\n

\u27f9 324 = 9 \u00d7 12d<\/p>\n\n\n\n

\u27f9 d = 3<\/p>\n\n\n\n

\u27f9 a = 2(3) [from (1)]<\/p>\n\n\n\n

\u27f9 a = 6<\/p>\n\n\n\n

Hence, the first term of the A.P. is 6<\/p>\n\n\n\n

For the 15th<\/sup> term, a15<\/sub> = a + 14d = 6 + 14 \u00d7 3 = 6 + 42<\/p>\n\n\n\n

Therefore, a15<\/sub> = 48<\/p>\n\n\n\n

29. If the 10th<\/sup> term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s consider a to be the first term and d be the common difference.<\/p>\n\n\n\n

And we know that, sum of first n terms is:<\/p>\n\n\n\n

Sn<\/sub> = n\/2(2a + (n \u2212 1)d) and nth<\/sup> term is given by: an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

Now, from the question we have<\/p>\n\n\n\n

S10<\/sub> = 120<\/p>\n\n\n\n

\u27f9 120 = 10\/2(2a + (10 \u2212 1)d)<\/p>\n\n\n\n

\u27f9 120 = 5(2a + 9d)<\/p>\n\n\n\n

\u27f9 24 = 2a + 9d \u2026. (1)<\/p>\n\n\n\n

Also given that, a10<\/sub> = 21<\/p>\n\n\n\n

\u27f9 21 = a + (10 \u2013 1)d<\/p>\n\n\n\n

\u27f9 21 = a + 9d \u2026. (2)<\/p>\n\n\n\n

Subtracting (2) from (1), we get<\/p>\n\n\n\n

24 \u2013 21 = 2a + 9d \u2013 a \u2013 9d<\/p>\n\n\n\n

\u27f9a = 3<\/p>\n\n\n\n

Now, on putting a = 3 in equation (2), we get<\/p>\n\n\n\n

3 + 9d = 21<\/p>\n\n\n\n

9d = 18<\/p>\n\n\n\n

d = 2<\/p>\n\n\n\n

Thus, we have the first term(a) = 3 and the common difference(d) = 2<\/p>\n\n\n\n

Therefore, the nth<\/sup> term is given by<\/p>\n\n\n\n

an<\/sub>  = a + (n \u2013 1)d = 3 + (n \u2013 1)2<\/p>\n\n\n\n

= 3 + 2n -2<\/p>\n\n\n\n

= 2n + 1<\/p>\n\n\n\n

Hence, the nth<\/sup> term of the A.P is (an<\/sub>) = 2n + 1.<\/p>\n\n\n\n

30. The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th<\/sup> term of this A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s take a to be the first term and d to be the common difference.<\/p>\n\n\n\n

And we know that, sum of first n terms<\/p>\n\n\n\n

Sn<\/sub> = n\/2(2a + (n \u2212 1)d)<\/p>\n\n\n\n

Given that sum of the first 7 terms of an A.P. is 63.<\/p>\n\n\n\n

S7<\/sub> = 63<\/p>\n\n\n\n

And sum of next 7 terms is 161.<\/p>\n\n\n\n

So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms<\/p>\n\n\n\n

S14 <\/sub>= 63 + 161 = 224<\/p>\n\n\n\n

Now, having<\/p>\n\n\n\n

S7<\/sub> = 7\/2(2a + (7 \u2212 1)d)<\/p>\n\n\n\n

\u27f9 63(2) = 7(2a + 6d)<\/p>\n\n\n\n

\u27f9 9 \u00d7 2 = 2a + 6d<\/p>\n\n\n\n

\u27f9 2a + 6d = 18 . . . . (1)<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

S14<\/sub> = 14\/2(2a + (14 \u2212 1)d)<\/p>\n\n\n\n

\u27f9 224 = 7(2a + 13d)<\/p>\n\n\n\n

\u27f9 32 = 2a + 13d \u2026. (2)<\/p>\n\n\n\n

Now, subtracting (1) from (2), we get<\/p>\n\n\n\n

\u27f9 13d \u2013 6d = 32 \u2013 18<\/p>\n\n\n\n

\u27f9 7d = 14<\/p>\n\n\n\n

\u27f9 d = 2<\/p>\n\n\n\n

Using d in (1), we have<\/p>\n\n\n\n

2a + 6(2) = 18<\/p>\n\n\n\n

2a = 18 \u2013 12<\/p>\n\n\n\n

a = 3<\/p>\n\n\n\n

Thus, from nth<\/sup> term<\/p>\n\n\n\n

\u27f9 a28  <\/sub>= a + (28 \u2013 1)d<\/p>\n\n\n\n

= 3 + 27 (2)<\/p>\n\n\n\n

= 3 + 54 = 57<\/p>\n\n\n\n

Therefore, the 28th<\/sup> term is 57.<\/p>\n\n\n\n

31. The sum of first seven terms of an A.P. is 182. If its 4th<\/sup> and 17th<\/sup> terms are in ratio 1: 5, find the A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that,<\/p>\n\n\n\n

S17<\/sub> = 182<\/p>\n\n\n\n

And, we know that the sum of first n term is:<\/p>\n\n\n\n

Sn <\/sub>= n\/2(2a + (n \u2212 1)d)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

S7<\/sub> = 7\/2(2a + (7 \u2212 1)d)<\/p>\n\n\n\n

182 \u00d7 2 = 7(2a + 6d)<\/p>\n\n\n\n

364 = 14a + 42d<\/p>\n\n\n\n

26 = a + 3d<\/p>\n\n\n\n

a = 26 \u2013 3d \u2026 (1)<\/p>\n\n\n\n

Also, it\u2019s given that 4th<\/sup> term and 17th<\/sup> term are in a ratio of 1: 5. So, we have<\/p>\n\n\n\n

\u27f9 5(a4<\/sub>) = 1(a17<\/sub>)<\/p>\n\n\n\n

\u27f9 5 (a + 3d) = 1 (a + 16d)<\/p>\n\n\n\n

\u27f9 5a + 15d = a + 16d<\/p>\n\n\n\n

\u27f9 4a = d \u2026. (2)<\/p>\n\n\n\n

Now, substituting (2) in (1), we get<\/p>\n\n\n\n

\u27f9 4 ( 26 \u2013 3d ) = d<\/p>\n\n\n\n

\u27f9 104 \u2013 12d = d<\/p>\n\n\n\n

\u27f9 104 = 13d<\/p>\n\n\n\n

\u27f9 d = 8<\/p>\n\n\n\n

Putting d in (2), we get<\/p>\n\n\n\n

\u27f9 4a = d<\/p>\n\n\n\n

\u27f9 4a = 8<\/p>\n\n\n\n

\u27f9 a = 2<\/p>\n\n\n\n

Therefore, the first term is 2 and the common difference is 8. So, the A.P. is 2, 10, 18, 26, . ..<\/p>\n\n\n\n

32. The nth<\/sup> term of an A.P is given by (-4n + 15). Find the sum of first 20 terms of this A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

The nth<\/sup> term of the A.P = (-4n + 15)<\/p>\n\n\n\n

So, by putting n = 1 and n = 20 we can find the first ans 20th<\/sup> term of the A.P<\/p>\n\n\n\n

a = (-4(1) + 15) = 11<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

a20<\/sub> = (-4(20) + 15) = -65<\/p>\n\n\n\n

Now, for find the sum of 20 terms of this A.P we have the first and last term.<\/p>\n\n\n\n

So, using the formula<\/p>\n\n\n\n

Sn<\/sub> = n\/2(a + l)<\/p>\n\n\n\n

S20<\/sub>  = 20\/2(11 + (-65))<\/p>\n\n\n\n

= 10(-54)<\/p>\n\n\n\n

= -540<\/p>\n\n\n\n

Therefore, the sum of first 20 terms of this A.P. is -540.<\/p>\n\n\n\n

33. In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let\u2019s take a to be the first term and d to be the common difference.<\/p>\n\n\n\n

And we know that, sum of first n terms<\/p>\n\n\n\n

Sn<\/sub> = n\/2(2a + (n \u2212 1)d)<\/p>\n\n\n\n

Given that sum of the first 10 terms of an A.P. is -150.<\/p>\n\n\n\n

S10<\/sub> = -150<\/p>\n\n\n\n

And the sum of next 10 terms is -550.<\/p>\n\n\n\n

So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms<\/p>\n\n\n\n

S20 <\/sub>= -150 + -550 = -700<\/p>\n\n\n\n

Now, having<\/p>\n\n\n\n

S10<\/sub> = 10\/2(2a + (10 \u2212 1)d)<\/p>\n\n\n\n

\u27f9 -150 = 5(2a + 9d)<\/p>\n\n\n\n

\u27f9 -30 = 2a + 9d<\/p>\n\n\n\n

\u27f9 2a + 9d = -30 . . . . (1)<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

S20<\/sub> = 20\/2(2a + (20 \u2212 1)d)<\/p>\n\n\n\n

\u27f9 -700 = 10(2a + 19d)<\/p>\n\n\n\n

\u27f9 -70 = 2a + 19d \u2026. (2)<\/p>\n\n\n\n

Now, subtracting (1) from (2), we get<\/p>\n\n\n\n

\u27f9 19d \u2013 9d = -70 \u2013 (-30)<\/p>\n\n\n\n

\u27f9 10d = -40<\/p>\n\n\n\n

\u27f9 d = -4<\/p>\n\n\n\n

Using d in (1), we have<\/p>\n\n\n\n

2a + 9(-4) = -30<\/p>\n\n\n\n

2a = -30 + 36<\/p>\n\n\n\n

a = 6\/2 = 3<\/p>\n\n\n\n

Hence, we have a = 3 and d = -4<\/p>\n\n\n\n

So, the A.P is 3, -1, -5, -9, -13,\u2026..<\/p>\n\n\n\n

34. Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th<\/sup> term.<\/strong><\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

First term of the A.P is 1505 and<\/p>\n\n\n\n

S14<\/sub> = 1505<\/p>\n\n\n\n

We know that, the sum of first n terms is<\/p>\n\n\n\n

Sn<\/sub> = n\/2(2a + (n \u2212 1)d)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

S14<\/sub> = 14\/2(2(10) + (14 \u2212 1)d) = 1505<\/p>\n\n\n\n

7(20 + 13d) = 1505<\/p>\n\n\n\n

20 + 13d = 215<\/p>\n\n\n\n

13d = 215 \u2013 20<\/p>\n\n\n\n

d = 195\/13<\/p>\n\n\n\n

d =15<\/p>\n\n\n\n

Thus, the 25th<\/sup> term is given by<\/p>\n\n\n\n

a25<\/sub> = 10 + (25 -1)15<\/p>\n\n\n\n

= 10 + (24)15<\/p>\n\n\n\n

= 10 + 360<\/p>\n\n\n\n

= 370<\/p>\n\n\n\n

Therefore, the 25th<\/sup> term of the A.P is 370<\/p>\n\n\n\n

35. In an A.P. , the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

The first term of the A.P. (a) = 2<\/p>\n\n\n\n

The last term of the A.P. (l) = 29<\/p>\n\n\n\n

And, sum of all the terms (Sn<\/sub>) = 155<\/p>\n\n\n\n

Let the common difference of the A.P. be d.<\/p>\n\n\n\n

So, find the number of terms by sum of terms formula<\/p>\n\n\n\n

Sn<\/sub> = n\/2 (a + l)<\/p>\n\n\n\n

155 = n\/2(2 + 29)<\/p>\n\n\n\n

155(2) = n(31)<\/p>\n\n\n\n

31n = 310<\/p>\n\n\n\n

n = 10<\/p>\n\n\n\n

Using n for the last term, we have<\/p>\n\n\n\n

l = a + (n \u2013 1)d<\/p>\n\n\n\n

29 = 2 + (10 \u2013 1)d<\/p>\n\n\n\n

29 = 2 + (9)d<\/p>\n\n\n\n

29 \u2013 2 = 9d<\/p>\n\n\n\n

9d = 27<\/p>\n\n\n\n

d = 3<\/p>\n\n\n\n

Hence, the common difference of the A.P. is d = 3<\/p>\n\n\n\n

36. The first and the last term of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

In an A.P first term (a) = 17 and the last term (l) = 350<\/p>\n\n\n\n

And, the common difference (d) = 9<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

so,<\/p>\n\n\n\n

an<\/sub> = l = 17 + (n \u2013 1)9 = 350<\/p>\n\n\n\n

17 + 9n \u2013 9 = 350<\/p>\n\n\n\n

9n = 350 \u2013 8<\/p>\n\n\n\n

n = 342\/9<\/p>\n\n\n\n

n = 38<\/p>\n\n\n\n

So, the sum of all the term of the A.P is given by<\/p>\n\n\n\n

Sn<\/sub> = n\/2 (a + l)<\/p>\n\n\n\n

= 38\/2(17 + 350)<\/p>\n\n\n\n

= 19(367)<\/p>\n\n\n\n

= 6973<\/p>\n\n\n\n

Therefore, the sum of terms of the A.P is 6973.<\/p>\n\n\n\n

37. Find the number of terms of the A.P. \u201312, \u20139, \u20136, . . . , 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

First term, a = -12<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1<\/sub> = \u2013 9 \u2013 (- 12)<\/p>\n\n\n\n

d = \u2013 9 + 12 = 3<\/p>\n\n\n\n

And, we know that nth<\/sup> term = an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

\u27f9 21 = -12 + (n \u2013 1)3<\/p>\n\n\n\n

\u27f9 21 = -12 + 3n \u2013 3<\/p>\n\n\n\n

\u27f9 21 = 3n \u2013 15<\/p>\n\n\n\n

\u27f9 36 = 3n<\/p>\n\n\n\n

\u27f9  n = 12<\/p>\n\n\n\n

Thus, the number of terms is 12.<\/p>\n\n\n\n

Now, if 1 is added to each of the 12 terms, the sum will increase by 12.<\/p>\n\n\n\n

Hence, the sum of all the terms of the A.P. so obtained is<\/p>\n\n\n\n

\u27f9 S12 <\/sub>+ 12  = 12\/2[a + l] + 12<\/p>\n\n\n\n

= 6[-12 + 21] + 12<\/p>\n\n\n\n

= 6 \u00d7 9 + 12<\/p>\n\n\n\n

= 66<\/p>\n\n\n\n

Therefore, the sum after adding 1 to each of the terms in the A.P is 66.<\/p>\n\n\n\n

RD Sharma Solutions for Class 10 Maths Chapter 9: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 10 Maths Chapter 9\u2013Arithmetic Progressions<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 10 Maths Chapter 9\u2013Arithmetic Progressions PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 10 Maths :<\/strong><\/h2>\n\n\n\n