{"id":543603,"date":"2021-09-30T06:39:08","date_gmt":"2021-09-30T06:39:08","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=543603"},"modified":"2021-10-01T09:04:27","modified_gmt":"2021-10-01T09:04:27","slug":"rd-sharma-solutions-for-class-11-maths-chapter-19-arithmetic-progressions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-19-arithmetic-progressions\/","title":{"rendered":"RD Sharma Solutions for Class 11 Maths Chapter 19\u2013Arithmetic Progressions"},"content":{"rendered":"\n

Class 11: Maths Chapter 19 solutions. Complete Class 11 Maths Chapter 19 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 19\u2013Arithmetic Progressions<\/h2>\n\n\n\n

RD Sharma 11th Maths Chapter 19, Class 11 Maths Chapter 19 solutions<\/p>\n\n\n\n

EXERCISE 19.1 PAGE NO: 19.4<\/p>\n\n\n\n

1. If the nth<\/sup> term of a sequence is given by an<\/sub> = n2<\/sup> \u2013 n+1, write down its first five terms.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

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

By using the values n = 1, 2, 3, 4, 5 we can find the first five terms.<\/p>\n\n\n\n

When n = 1:<\/p>\n\n\n\n

a1<\/sub> = (1)2<\/sup> \u2013 1 + 1<\/p>\n\n\n\n

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

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

When n = 2:<\/p>\n\n\n\n

a2<\/sub> = (2)2<\/sup> \u2013 2 + 1<\/p>\n\n\n\n

= 4 \u2013 2 + 1<\/p>\n\n\n\n

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

When n = 3:<\/p>\n\n\n\n

a3<\/sub> = (3)2<\/sup> \u2013 3 + 1<\/p>\n\n\n\n

= 9 \u2013 3 + 1<\/p>\n\n\n\n

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

When n = 4:<\/p>\n\n\n\n

a4<\/sub> = (4)2<\/sup> \u2013 4 + 1<\/p>\n\n\n\n

= 16 \u2013 4 + 1<\/p>\n\n\n\n

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

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

a5<\/sub> = (5)2<\/sup> \u2013 5 + 1<\/p>\n\n\n\n

= 25 \u2013 5 + 1<\/p>\n\n\n\n

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

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

2. A sequence is defined by an<\/sub> = n3<\/sup> \u2013 6n2<\/sup> + 11n \u2013 6, n \u2208 N. Show that the first three terms of the sequence are zero and all other terms are positive.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

an<\/sub> = n3<\/sup> \u2013 6n2<\/sup> + 11n \u2013 6, n \u2208 N<\/p>\n\n\n\n

By using the values n = 1, 2, 3 we can find the first three terms.<\/p>\n\n\n\n

When n = 1:<\/p>\n\n\n\n

a1<\/sub> = (1)3<\/sup> \u2013 6(1)2<\/sup> + 11(1) \u2013 6<\/p>\n\n\n\n

= 1 \u2013 6 + 11 \u2013 6<\/p>\n\n\n\n

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

= 0<\/p>\n\n\n\n

When n = 2:<\/p>\n\n\n\n

a2<\/sub> = (2)3<\/sup> \u2013 6(2)2<\/sup> + 11(2) \u2013 6<\/p>\n\n\n\n

= 8 \u2013 6(4) + 22 \u2013 6<\/p>\n\n\n\n

= 8 \u2013 24 + 22 \u2013 6<\/p>\n\n\n\n

= 30 \u2013 30<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

When n = 3:<\/p>\n\n\n\n

a3<\/sub> = (3)3<\/sup> \u2013 6(3)2<\/sup> + 11(3) \u2013 6<\/p>\n\n\n\n

= 27 \u2013 6(9) + 33 \u2013 6<\/p>\n\n\n\n

= 27 \u2013 54 + 33 \u2013 6<\/p>\n\n\n\n

= 60 \u2013 60<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

This shows that the first three terms of the sequence is zero.<\/p>\n\n\n\n

Now, let\u2019s check for when n = n:<\/p>\n\n\n\n

an<\/sub> = n3<\/sup> \u2013 6n2<\/sup> + 11n \u2013 6<\/p>\n\n\n\n

= n3<\/sup> \u2013 6n2<\/sup> + 11n \u2013 6 \u2013 n + n \u2013 2 + 2<\/p>\n\n\n\n

= n3<\/sup> \u2013 6n2<\/sup> + 12n \u2013 8 \u2013 n + 2<\/p>\n\n\n\n

= (n)3<\/sup> \u2013 3\u00d72n(n \u2013 2) \u2013 (2)3<\/sup> \u2013 n + 2<\/p>\n\n\n\n

By using the formula, {(a \u2013 b)3<\/sup> = (a)3<\/sup> \u2013 (b)3<\/sup> \u2013 3ab(a \u2013 b)}<\/p>\n\n\n\n

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

Here, n \u2013 2 will always be positive for n > 3<\/p>\n\n\n\n

\u2234 an<\/sub> is always positive for n > 3<\/p>\n\n\n\n

3. Find the first four terms of the sequence defined by a1<\/sub> = 3 and an<\/sub> = 3an\u20131<\/sub> + 2, for all n > 1.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

a1<\/sub> = 3 and an<\/sub> = 3an\u20131<\/sub> + 2, for all n > 1<\/p>\n\n\n\n

By using the values n = 1, 2, 3, 4 we can find the first four terms.<\/p>\n\n\n\n

When n = 1:<\/p>\n\n\n\n

a1<\/sub> = 3<\/p>\n\n\n\n

When n = 2:<\/p>\n\n\n\n

a2<\/sub> = 3a2\u20131<\/sub> + 2<\/p>\n\n\n\n

= 3a1<\/sub> + 2<\/p>\n\n\n\n

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

= 9 + 2<\/p>\n\n\n\n

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

When n = 3:<\/p>\n\n\n\n

a3<\/sub> = 3a3\u20131<\/sub> + 2<\/p>\n\n\n\n

= 3a2<\/sub> + 2<\/p>\n\n\n\n

= 3(11) + 2<\/p>\n\n\n\n

= 33 + 2<\/p>\n\n\n\n

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

When n = 4:<\/p>\n\n\n\n

a4<\/sub> = 3a4\u20131<\/sub> + 2<\/p>\n\n\n\n

= 3a3<\/sub> + 2<\/p>\n\n\n\n

= 3(35) + 2<\/p>\n\n\n\n

= 105 + 2<\/p>\n\n\n\n

= 107<\/p>\n\n\n\n

\u2234 First four terms of sequence are 3, 11, 35, 107.<\/p>\n\n\n\n

4. Write the first five terms in each of the following sequences:
(i) a1<\/sub> = 1, an<\/sub> = an\u20131<\/sub> + 2, n > 1<\/strong><\/p>\n\n\n\n

(ii) a1<\/sub> = 1 = a2<\/sub>, an<\/sub> = an\u20131<\/sub> + an\u20132<\/sub>, n > 2<\/strong><\/p>\n\n\n\n

(iii) a1<\/sub> = a2<\/sub> =2, an<\/sub> = an\u20131<\/sub> \u2013 1, n > 2<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>a1<\/sub> = 1, an<\/sub> = an\u20131<\/sub> + 2, n > 1<\/p>\n\n\n\n

By using the values n = 1, 2, 3, 4, 5 we can find the first five terms.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a1<\/sub> = 1<\/p>\n\n\n\n

When n = 2:<\/p>\n\n\n\n

a2<\/sub> = a2\u20131<\/sub> + 2<\/p>\n\n\n\n

= a1<\/sub> + 2<\/p>\n\n\n\n

= 1 + 2<\/p>\n\n\n\n

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

When n = 3:<\/p>\n\n\n\n

a3<\/sub> = a3\u20131<\/sub> + 2<\/p>\n\n\n\n

= a2<\/sub> + 2<\/p>\n\n\n\n

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

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

When n = 4:<\/p>\n\n\n\n

a4<\/sub> = a4\u20131<\/sub> + 2<\/p>\n\n\n\n

= a3<\/sub> + 2<\/p>\n\n\n\n

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

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

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

a5<\/sub> = a5\u20131<\/sub> + 2<\/p>\n\n\n\n

= a4<\/sub> + 2<\/p>\n\n\n\n

= 7 + 2<\/p>\n\n\n\n

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

\u2234 First five terms of the sequence are 1, 3, 5, 7, 9.<\/p>\n\n\n\n

(ii) <\/strong>a1<\/sub> = 1 = a2<\/sub>, an<\/sub> = an\u20131<\/sub> + an\u20132<\/sub>, n > 2<\/p>\n\n\n\n

By using the values n = 1, 2, 3, 4, 5 we can find the first five terms.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a1<\/sub> = 1<\/p>\n\n\n\n

a2<\/sub> = 1<\/p>\n\n\n\n

When n = 3:<\/p>\n\n\n\n

a3<\/sub> = a3\u20131<\/sub> + a3\u20132<\/sub><\/p>\n\n\n\n

= a2<\/sub> + a1<\/sub><\/p>\n\n\n\n

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

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

When n = 4:<\/p>\n\n\n\n

a4<\/sub> = a4\u20131<\/sub> + a4\u20132<\/sub><\/p>\n\n\n\n

= a3<\/sub> + a2<\/sub><\/p>\n\n\n\n

= 2 + 1<\/p>\n\n\n\n

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

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

a5<\/sub> = a5\u20131<\/sub> + a5\u20132<\/sub><\/p>\n\n\n\n

= a4<\/sub> + a3<\/sub><\/p>\n\n\n\n

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

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

\u2234 First five terms of the sequence are 1, 1, 2, 3, 5.<\/p>\n\n\n\n

(iii) <\/strong>a1<\/sub> = a2<\/sub> =2, an<\/sub> = an\u20131<\/sub> \u2013 1, n > 2<\/p>\n\n\n\n

By using the values n = 1, 2, 3, 4, 5 we can find the first five terms.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a1<\/sub> = 2<\/p>\n\n\n\n

a2<\/sub> = 2<\/p>\n\n\n\n

When n = 3:<\/p>\n\n\n\n

a3<\/sub> = a3\u20131<\/sub> \u2013 1<\/p>\n\n\n\n

= a2<\/sub> \u2013 1<\/p>\n\n\n\n

= 2 \u2013 1<\/p>\n\n\n\n

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

When n = 4:<\/p>\n\n\n\n

a4<\/sub> = a4\u20131<\/sub> \u2013 1<\/p>\n\n\n\n

= a3<\/sub> \u2013 1<\/p>\n\n\n\n

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

= 0<\/p>\n\n\n\n

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

a5<\/sub> = a5\u20131<\/sub> \u2013 1<\/p>\n\n\n\n

= a4<\/sub> \u2013 1<\/p>\n\n\n\n

= 0 \u2013 1<\/p>\n\n\n\n

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

\u2234 First five terms of the sequence are 2, 2, 1, 0, -1.<\/p>\n\n\n\n

5. The Fibonacci sequence is defined by a1<\/sub> = 1 = a2<\/sub>, an<\/sub> = an\u20131<\/sub> + an\u20132 <\/sub>for n > 2. Find (an+1<\/sub>)\/an<\/sub> for n = 1, 2, 3, 4, 5.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

a1<\/sub> = 1<\/p>\n\n\n\n

a2<\/sub> = 1<\/p>\n\n\n\n

an<\/sub> = an\u20131<\/sub> + an\u20132<\/sub><\/p>\n\n\n\n

When n = 1:<\/p>\n\n\n\n

(an+1<\/sub>)\/an <\/sub>= (a1+1<\/sub>)\/a1<\/sub><\/p>\n\n\n\n

= a2<\/sub>\/a1<\/sub><\/p>\n\n\n\n

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

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

a3<\/sub> = a3\u20131<\/sub> + a3\u20132<\/sub><\/p>\n\n\n\n

= a2<\/sub> + a1<\/sub><\/p>\n\n\n\n

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

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

When n = 2:<\/p>\n\n\n\n

(an+1<\/sub>)\/an <\/sub>= (a2+1<\/sub>)\/a2<\/sub><\/p>\n\n\n\n

= a3<\/sub>\/a2<\/sub><\/p>\n\n\n\n

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

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

a4<\/sub> = a4\u20131<\/sub> + a4\u20132<\/sub><\/p>\n\n\n\n

= a3<\/sub> + a2<\/sub><\/p>\n\n\n\n

= 2 + 1<\/p>\n\n\n\n

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

When n = 3:<\/p>\n\n\n\n

(an+1<\/sub>)\/an <\/sub>= (a3+1<\/sub>)\/a3<\/sub><\/p>\n\n\n\n

= a4<\/sub>\/a3<\/sub><\/p>\n\n\n\n

= 3\/2<\/p>\n\n\n\n

a5<\/sub> = a5\u20131<\/sub> + a5\u20132<\/sub><\/p>\n\n\n\n

= a4<\/sub> + a3<\/sub><\/p>\n\n\n\n

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

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

When n = 4:<\/p>\n\n\n\n

(an+1<\/sub>)\/an <\/sub>= (a4+1<\/sub>)\/a4<\/sub><\/p>\n\n\n\n

= a5<\/sub>\/a4<\/sub><\/p>\n\n\n\n

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

a6<\/sub> = a6\u20131<\/sub> + a6\u20132<\/sub><\/p>\n\n\n\n

= a5<\/sub> + a4<\/sub><\/p>\n\n\n\n

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

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

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

(an+1<\/sub>)\/an <\/sub>= (a5+1<\/sub>)\/a5<\/sub><\/p>\n\n\n\n

= a6<\/sub>\/a5<\/sub> = 8\/5<\/p>\n\n\n\n

\u2234 Value of (an+1<\/sub>)\/an<\/sub> when n = 1, 2, 3, 4, 5 are 1, 2, 3\/2, 5\/3, 8\/5<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 19.2 PAGE NO: 19.11<\/p>\n\n\n\n

1. Find:
(i) 10th<\/sup> term of the A.P. 1, 4, 7, 10, \u2026..<\/strong><\/p>\n\n\n\n

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

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

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

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

Arithmetic Progression (AP) whose common difference is = an<\/sub> \u2013 an-1<\/sub> where n > 0<\/p>\n\n\n\n

Let us consider, a = a1<\/sub> = 1, a2<\/sub> = 4 \u2026<\/p>\n\n\n\n

So, Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 4 \u2013 1 = 3<\/p>\n\n\n\n

To find the 10th<\/sup> term of A.P, firstly find an<\/sub><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

an<\/sub> = a + (n-1) d<\/p>\n\n\n\n

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

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

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

When n = 10:<\/p>\n\n\n\n

a10<\/sub> = 3(10) \u2013 2<\/p>\n\n\n\n

= 30 \u2013 2<\/p>\n\n\n\n

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

Hence, 10th<\/sup> term is 28.<\/p>\n\n\n\n

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

Arithmetic Progression (AP) whose common difference is = an<\/sub> \u2013 an-1<\/sub> where n > 0<\/p>\n\n\n\n

Let us consider, a = a1<\/sub> = \u221a2, a2<\/sub> = 3\u221a2 \u2026<\/p>\n\n\n\n

So, Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 3\u221a2 \u2013 \u221a2 = 2\u221a2<\/p>\n\n\n\n

To find the 18th<\/sup> term of A.P, firstly find an<\/sub><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

an<\/sub> = a + (n-1) d<\/p>\n\n\n\n

= \u221a2 + (n \u2013 1) 2\u221a2<\/p>\n\n\n\n

= \u221a2 + 2\u221a2n \u2013 2\u221a2<\/p>\n\n\n\n

= 2\u221a2n \u2013 \u221a2<\/p>\n\n\n\n

When n = 18:<\/p>\n\n\n\n

a18<\/sub> = 2\u221a2(18) \u2013 \u221a2<\/p>\n\n\n\n

= 36\u221a2 \u2013 \u221a2<\/p>\n\n\n\n

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

Hence, 10th<\/sup> term is 35\u221a2<\/p>\n\n\n\n

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

Arithmetic Progression (AP) whose common difference is = an<\/sub> \u2013 an-1<\/sub> where n > 0<\/p>\n\n\n\n

Let us consider, a = a1<\/sub> = 13, a2<\/sub> = 8 \u2026<\/p>\n\n\n\n

So, Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 8 \u2013 13 = -5<\/p>\n\n\n\n

To find the nth<\/sup> term of A.P, firstly find an<\/sub><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

an<\/sub> = a + (n-1) d<\/p>\n\n\n\n

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

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

= 18 \u2013 5n<\/p>\n\n\n\n

Hence, nth<\/sup> term is 18 \u2013 5n<\/p>\n\n\n\n

2. In an A.P., show that am+n<\/sub> + am\u2013n<\/sub> = 2am<\/sub>.<\/strong><\/p>\n\n\n\n

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

We know the first term is \u2018a\u2019 and the common difference of an A.P is d.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

am+n<\/sub> + am\u2013n<\/sub> = 2am<\/sub><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

Now, let us take LHS: am+n<\/sub> + am-n<\/sub><\/p>\n\n\n\n

am+n<\/sub> + am-n<\/sub> = a + (m + n \u2013 1)d + a + (m \u2013 n \u2013 1)d<\/p>\n\n\n\n

= a + md + nd \u2013 d + a + md \u2013 nd \u2013 d<\/p>\n\n\n\n

= 2a + 2md \u2013 2d<\/p>\n\n\n\n

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

= 2[a + d(m \u2013 1)] {\u2235 an<\/sub> = a + (n \u2013 1)d}<\/p>\n\n\n\n

am+n<\/sub> + am-n<\/sub> = 2am<\/sub><\/p>\n\n\n\n

Hence Proved.<\/p>\n\n\n\n

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

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

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

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

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

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

Here, a1<\/sub> = a = 3, a2<\/sub> = 8<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 8 \u2013 3 = 5<\/p>\n\n\n\n

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

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

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

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

Now, to find which term of A.P is 248<\/p>\n\n\n\n

Put an<\/sub> = 248<\/p>\n\n\n\n

\u2234 5n \u2013 2 = 248<\/p>\n\n\n\n

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

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

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

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

Hence, 50th<\/sup> term of given A.P is 248.<\/p>\n\n\n\n

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

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

Here, a1<\/sub> = a = 84, a2<\/sub> = 88<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 80 \u2013 84 = -4<\/p>\n\n\n\n

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

an<\/sub> = 84 + (n \u2013 1)-4<\/p>\n\n\n\n

= 84 \u2013 4n + 4<\/p>\n\n\n\n

= 88 \u2013 4n<\/p>\n\n\n\n

Now, to find which term of A.P is 0<\/p>\n\n\n\n

Put an<\/sub> = 0<\/p>\n\n\n\n

88 \u2013 4n = 0<\/p>\n\n\n\n

-4n = -88<\/p>\n\n\n\n

n = 88\/4<\/p>\n\n\n\n

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

Hence, 22nd<\/sup> term of given A.P is 0.<\/p>\n\n\n\n

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

Given A.P is 4, 9, 14,\u2026<\/p>\n\n\n\n

Here, a1<\/sub> = a = 4, a2<\/sub> = 9<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 9 \u2013 4 = 5<\/p>\n\n\n\n

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

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

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

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

Now, to find which term of A.P is 254<\/p>\n\n\n\n

Put an<\/sub> = 254<\/p>\n\n\n\n

5n \u2013 1 = 254<\/p>\n\n\n\n

5n = 254 + 1<\/p>\n\n\n\n

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

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

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

Hence, 51st<\/sup> term of given A.P is 254.<\/p>\n\n\n\n

4. (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

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

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

Given A.P is 7, 10, 13,\u2026<\/p>\n\n\n\n

Here, a1<\/sub> = a = 7, a2<\/sub> = 10<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 10 \u2013 7 = 3<\/p>\n\n\n\n

We know, an<\/sub> = a + (n \u2013 1)d [where, a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

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

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

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

Now, to find whether 68 is a term of this A.P. or not<\/p>\n\n\n\n

Put an<\/sub> = 68<\/p>\n\n\n\n

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

3n = 68 \u2013 4<\/p>\n\n\n\n

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

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

64\/3 is not a natural number<\/p>\n\n\n\n

Hence, 68 is not a term of given A.P.<\/p>\n\n\n\n

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

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

Here, a1<\/sub> = a = 3, a2<\/sub> = 8<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 8 \u2013 3 = 5<\/p>\n\n\n\n

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

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

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

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

To find whether 302 is a term of this A.P. or not<\/p>\n\n\n\n

Put an<\/sub> = 302<\/p>\n\n\n\n

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

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

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

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

304\/5 is not a natural number<\/p>\n\n\n\n

Hence, 304 is not a term of given A.P.<\/p>\n\n\n\n

5. (i) Which term of the sequence 24, 23 \u00bc, 22 \u00bd, 21 \u00be is the first negative term?<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

AP: 24, 23 \u00bc, 22 \u00bd, 21 \u00be, \u2026 = 24, 93\/4, 45\/2, 87\/4, \u2026<\/p>\n\n\n\n

Here, a1<\/sub> = a = 24, a2<\/sub> = 93\/4<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 93\/4 \u2013 24<\/p>\n\n\n\n

= (93 \u2013 96)\/4<\/p>\n\n\n\n

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

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

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

an<\/sub> = 24 + (n \u2013 1) (-3\/4)<\/p>\n\n\n\n

= 24 \u2013 3\/4n + \u00be<\/p>\n\n\n\n

= (96+3)\/4 \u2013 3\/4n<\/p>\n\n\n\n

= 99\/4 \u2013 3\/4n<\/p>\n\n\n\n

Now we need to find, first negative term.<\/p>\n\n\n\n

Put an<\/sub> < 0<\/p>\n\n\n\n

an<\/sub> = 99\/4 \u2013 3\/4n < 0<\/p>\n\n\n\n

99\/4 < 3\/4n<\/p>\n\n\n\n

3n > 99<\/p>\n\n\n\n

n > 99\/3<\/p>\n\n\n\n

n > 33<\/p>\n\n\n\n

Hence, 34th<\/sup> term is the first negative term of given AP.<\/p>\n\n\n\n

(ii) Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, \u2026 is (a) purely real (b) purely imaginary ?<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

AP: 12 + 8i, 11 + 6i, 10 + 4i, \u2026<\/p>\n\n\n\n

Here, a1<\/sub> = a = 12 + 8i, a2<\/sub> = 11 + 6i<\/p>\n\n\n\n

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

= 11 + 6i \u2013 (12 + 8i)<\/p>\n\n\n\n

= 11 \u2013 12 + 6i \u2013 8i<\/p>\n\n\n\n

= -1 \u2013 2i<\/p>\n\n\n\n

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

an<\/sub> = 12 + 8i + (n \u2013 1) -1 \u2013 2i<\/p>\n\n\n\n

= 12 + 8i \u2013 n \u2013 2ni + 1 + 2i<\/p>\n\n\n\n

= 13 + 10i \u2013 n \u2013 2ni<\/p>\n\n\n\n

= (13 \u2013 n) + (10 \u2013 2n) i<\/p>\n\n\n\n

To find purely real term of this A.P., imaginary part have to be zero<\/p>\n\n\n\n

10 \u2013 2n = 0<\/p>\n\n\n\n

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

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

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

Hence, 5th<\/sup> term is purely real.<\/p>\n\n\n\n

To find purely imaginary term of this A.P., real part have to be zero<\/p>\n\n\n\n

\u2234 13 \u2013 n = 0<\/p>\n\n\n\n

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

Hence, 13th<\/sup> term is purely imaginary.<\/p>\n\n\n\n

6. (i) How many terms are in A.P. 7, 10, 13,\u202643?<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

AP: 7, 10, 13,\u2026<\/p>\n\n\n\n

Here, a1<\/sub> = a = 7, a2<\/sub> = 10<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1<\/sub> = 10 \u2013 7 = 3<\/p>\n\n\n\n

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

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

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

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

To find total terms of the A.P., put an<\/sub> = 43 as 43 is last term of A.P.<\/p>\n\n\n\n

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

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

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

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

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

Hence, total 13 terms exists in the given A.P.<\/p>\n\n\n\n

(ii) How many terms are there in the A.P. -1, -5\/6, -2\/3, -1\/2, \u2026, 10\/3 ?<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

AP: -1, -5\/6, -2\/3, -1\/2, \u2026<\/p>\n\n\n\n

Here, a1<\/sub> = a = -1, a2<\/sub> = -5\/6<\/p>\n\n\n\n

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

= -5\/6 \u2013 (-1)<\/p>\n\n\n\n

= -5\/6 + 1<\/p>\n\n\n\n

= (-5+6)\/6<\/p>\n\n\n\n

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

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

an<\/sub> = -1 + (n \u2013 1) 1\/6<\/p>\n\n\n\n

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

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

= -7\/6 + 1\/6n<\/p>\n\n\n\n

To find total terms of the AP,<\/p>\n\n\n\n

Put an<\/sub> = 10\/3 [Since, 10\/3 is the last term of AP]<\/p>\n\n\n\n

an<\/sub> = -7\/6 + 1\/6n = 10\/3<\/p>\n\n\n\n

1\/6n = 10\/3 + 7\/6<\/p>\n\n\n\n

1\/6n = (20+7)\/6<\/p>\n\n\n\n

1\/6n = 27\/6<\/p>\n\n\n\n

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

Hence, total 27 terms exists in the given A.P.<\/p>\n\n\n\n

7. 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

First term, a = 5; last term, l = an<\/sub> = 80<\/p>\n\n\n\n

Common difference, d = 3<\/p>\n\n\n\n

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

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

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

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

To find total terms of the A.P., put an<\/sub> = 80 as 80 is last term of A.P.<\/p>\n\n\n\n

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

3n = 80 \u2013 2<\/p>\n\n\n\n

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

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

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

Hence, total 26 terms exists in the given A.P.<\/p>\n\n\n\n

8. 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

6th<\/sup> term of an A.P is 19 and 17th<\/sup> terms of an A.P. is 41<\/p>\n\n\n\n

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

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

When n = 6:<\/p>\n\n\n\n

a6<\/sub> = a + (6 \u2013 1) d<\/p>\n\n\n\n

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

Similarly, When n = 17:<\/p>\n\n\n\n

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

= a + 16d<\/p>\n\n\n\n

According to question:<\/p>\n\n\n\n

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

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

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

Let us subtract equation (i) from (ii) we get,<\/p>\n\n\n\n

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

a + 16d \u2013 a \u2013 5d = 22<\/p>\n\n\n\n

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

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

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

put the value of d in equation (i):<\/p>\n\n\n\n

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

a + 10 = 19<\/p>\n\n\n\n

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

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

As, an<\/sub> = a + (n \u2013 1)d<\/p>\n\n\n\n

a40<\/sub> = a + (40 \u2013 1)d<\/p>\n\n\n\n

= a + 39d<\/p>\n\n\n\n

Now put the value of a = 9 and d = 2 in a40<\/sub> we get,<\/p>\n\n\n\n

a40<\/sub> = 9 + 39(2)<\/p>\n\n\n\n

= 9 + 78<\/p>\n\n\n\n

= 87<\/p>\n\n\n\n

Hence, 40th<\/sup> term of the given A.P. is 87.<\/p>\n\n\n\n

9. If 9th<\/sup> term of an A.P. is Zero, prove that 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

9th<\/sup> term of an A.P is 0<\/p>\n\n\n\n

So, a9<\/sub> = 0<\/p>\n\n\n\n

We need to prove: a29<\/sub> = 2a19<\/sub><\/p>\n\n\n\n

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

When n = 9:<\/p>\n\n\n\n

a9<\/sub> = a + (9 \u2013 1)d<\/p>\n\n\n\n

= a + 8d<\/p>\n\n\n\n

According to question:<\/p>\n\n\n\n

a9<\/sub> = 0<\/p>\n\n\n\n

a + 8d = 0<\/p>\n\n\n\n

a = -8d<\/p>\n\n\n\n

When n = 19:<\/p>\n\n\n\n

a19<\/sub> = a + (19 \u2013 1)d<\/p>\n\n\n\n

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

= -8d + 18d<\/p>\n\n\n\n

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

When n = 29:<\/p>\n\n\n\n

a29<\/sub> = a + (29 \u2013 1)d<\/p>\n\n\n\n

= a + 28d<\/p>\n\n\n\n

= -8d + 28d [Since, a = -8d]<\/p>\n\n\n\n

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

= 2\u00d710d<\/p>\n\n\n\n

a29<\/sub> = 2a19<\/sub> [Since, a19<\/sub> = 10d]<\/p>\n\n\n\n

Hence Proved.<\/p>\n\n\n\n

10. If 10 times the 10th<\/sup> term of an A.P. is equal to 15 times the 15th<\/sup> term, show that the 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

So, 10a10<\/sub> = 15a15<\/sub><\/p>\n\n\n\n

We need to prove: a25<\/sub> = 0<\/p>\n\n\n\n

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

When n = 10:<\/p>\n\n\n\n

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

= a + 9d<\/p>\n\n\n\n

When n = 15:<\/p>\n\n\n\n

a15<\/sub> = a + (15 \u2013 1)d<\/p>\n\n\n\n

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

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

a25<\/sub> = a + (25 \u2013 1)d<\/p>\n\n\n\n

= a + 24d \u2026\u2026\u2026(i)<\/p>\n\n\n\n

According to question:<\/p>\n\n\n\n

10a10<\/sub> = 15a15<\/sub><\/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

10a \u2013 15a + 90d \u2013 210d = 0<\/p>\n\n\n\n

-5a \u2013 120d = 0<\/p>\n\n\n\n

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

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

a25<\/sub> = 0 [From (i)]<\/p>\n\n\n\n

Hence Proved.<\/p>\n\n\n\n

11. The 10th<\/sup> and 18th<\/sup> term 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

10th<\/sup> term of an A.P is 41, and 18th<\/sup> terms of an A.P. is 73<\/p>\n\n\n\n

So, a10<\/sub> = 41 and a18<\/sub> = 73<\/p>\n\n\n\n

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is the common difference and n is any natural number]<\/p>\n\n\n\n

When n = 10:<\/p>\n\n\n\n

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

= a + 9d<\/p>\n\n\n\n

When n = 18:<\/p>\n\n\n\n

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

= a + 17d<\/p>\n\n\n\n

According to question:<\/p>\n\n\n\n

a10<\/sub> = 41 and a18<\/sub> = 73<\/p>\n\n\n\n

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

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

Let us subtract equation (i) from (ii) we get,<\/p>\n\n\n\n

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

a + 17d \u2013 a \u2013 9d = 32<\/p>\n\n\n\n

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

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

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

Put the value of d in equation (i) we get,<\/p>\n\n\n\n

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

a + 36 = 41<\/p>\n\n\n\n

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

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

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

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

= a + 25d<\/p>\n\n\n\n

Now put the value of a = 5 and d = 4 in a26<\/sub><\/p>\n\n\n\n

a26<\/sub> = 5 + 25(4)<\/p>\n\n\n\n

= 5 + 100<\/p>\n\n\n\n

= 105<\/p>\n\n\n\n

Hence, 26th<\/sup> term of the given A.P. is 105.<\/p>\n\n\n\n

12. 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

So, a24<\/sub> = 2a10<\/sub><\/p>\n\n\n\n

We need to prove: a72<\/sub> = 2a34<\/sub><\/p>\n\n\n\n

We know, an<\/sub> = a + (n \u2013 1) d [where a is first term or a1<\/sub> and d is common difference and n is any natural number]<\/p>\n\n\n\n

When n = 10:<\/p>\n\n\n\n

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

= a + 9d<\/p>\n\n\n\n

When n = 24:<\/p>\n\n\n\n

a24<\/sub> = a + (24 \u2013 1)d<\/p>\n\n\n\n

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

When n = 34:<\/p>\n\n\n\n

a34<\/sub> = a + (34 \u2013 1)d<\/p>\n\n\n\n

= a + 33d \u2026\u2026\u2026(i)<\/p>\n\n\n\n

When n = 72:<\/p>\n\n\n\n

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

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

According to question:<\/p>\n\n\n\n

a24<\/sub> = 2a10<\/sub><\/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 \u2013 2a + 23d \u2013 18d = 0<\/p>\n\n\n\n

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

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

Now, a72<\/sub> = a + 71d<\/p>\n\n\n\n

a72<\/sub> = 5d + 71d<\/p>\n\n\n\n

= 76d<\/p>\n\n\n\n

= 10d + 66d<\/p>\n\n\n\n

= 2(5d + 33d)<\/p>\n\n\n\n

= 2(a + 33d) [since, a = 5d]<\/p>\n\n\n\n

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

Hence Proved.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 19.3 PAGE NO: 19.15<\/p>\n\n\n\n

1. The Sum of the three terms of an A.P. is 21 and the product of the first, and the third terms exceed the second term by 6, find three terms.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

The sum of first three terms is 21<\/p>\n\n\n\n

Let us assume the first three terms as a \u2013 d, a, a + d [where a is the first term and d is the common difference]<\/p>\n\n\n\n

So, sum of first three terms is<\/p>\n\n\n\n

a \u2013 d + a + a + d = 21<\/p>\n\n\n\n

3a = 21<\/p>\n\n\n\n

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

It is also given that product of first and third term exceeds the second by 6<\/p>\n\n\n\n

So, (a \u2013 d)(a + d) \u2013 a = 6<\/p>\n\n\n\n

a2<\/sup> \u2013 d2<\/sup> \u2013 a = 6<\/p>\n\n\n\n

Substituting the value of a = 7, we get<\/p>\n\n\n\n

72<\/sup> \u2013 d2<\/sup> \u2013 7 = 6<\/p>\n\n\n\n

d2<\/sup> = 36<\/p>\n\n\n\n

d = 6 or d = \u2013 6<\/p>\n\n\n\n

Hence, the terms of AP are a \u2013 d, a, a + d which is 1, 7, 13.<\/p>\n\n\n\n

2. Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

Sum of first three terms is 27<\/p>\n\n\n\n

Let us assume the first three terms as a \u2013 d, a, a + d [where a is the first term and d is the common difference]<\/p>\n\n\n\n

So, sum of first three terms is<\/p>\n\n\n\n

a \u2013 d + a + a + d = 27<\/p>\n\n\n\n

3a = 27<\/p>\n\n\n\n

a = 9<\/p>\n\n\n\n

It is given that the product of three terms is 648<\/p>\n\n\n\n

So, a3<\/sup> \u2013 ad2<\/sup> = 648<\/p>\n\n\n\n

Substituting the value of a = 9, we get<\/p>\n\n\n\n

93<\/sup> \u2013 9d2<\/sup> = 648<\/p>\n\n\n\n

729 \u2013 9d2<\/sup> = 648<\/p>\n\n\n\n

81 = 9d2<\/sup><\/p>\n\n\n\n

d = 3 or d = \u2013 3<\/p>\n\n\n\n

Hence, the given terms are a \u2013 d, a, a + d which is 6, 9, 12.<\/p>\n\n\n\n

3. Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

Sum of four terms is 50.<\/p>\n\n\n\n

Let us assume these four terms as a \u2013 3d, a \u2013 d, a + d, a + 3d<\/p>\n\n\n\n

It is given that, sum of these terms is 4a = 50<\/p>\n\n\n\n

So, a = 50\/4<\/p>\n\n\n\n

= 25\/2 \u2026 (i)<\/p>\n\n\n\n

It is also given that the greatest number is 4 time the least<\/p>\n\n\n\n

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

Substitute the value of a = 25\/2, we get<\/p>\n\n\n\n

(25+6d)\/2 = 50 \u2013 12d<\/p>\n\n\n\n

30d = 75<\/p>\n\n\n\n

d = 75\/30<\/p>\n\n\n\n

= 25\/10<\/p>\n\n\n\n

= 5\/2 \u2026 (ii)<\/p>\n\n\n\n

Hence, the terms of AP are a \u2013 3d, a \u2013 d, a + d, a + 3d which is 5, 10, 15, 20<\/p>\n\n\n\n

4. The sum of three numbers in A.P. is 12, and the sum of their cubes is 288. Find the numbers.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

The sum of three numbers is 12<\/p>\n\n\n\n

Let us assume the numbers in AP are a \u2013 d, a, a + d<\/p>\n\n\n\n

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

3a = 12<\/p>\n\n\n\n

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

It is also given that the sum of their cube is 288<\/p>\n\n\n\n

(a \u2013 d)3<\/sup> + a3<\/sup> + (a + d)3<\/sup> = 288<\/p>\n\n\n\n

a3<\/sup> \u2013 d3<\/sup> \u2013 3ad(a \u2013 d) + a3<\/sup> + a3<\/sup> + d3<\/sup> + 3ad(a + d) = 288<\/p>\n\n\n\n

Substitute the value of a = 4, we get<\/p>\n\n\n\n

64 \u2013 d3<\/sup> \u2013 12d(4 \u2013 d) + 64 + 64 + d3<\/sup> + 12d(4 + d) = 288<\/p>\n\n\n\n

192 + 24d2<\/sup> = 288<\/p>\n\n\n\n

d = 2 or d = \u2013 2<\/p>\n\n\n\n

Hence, the numbers are a \u2013 d, a, a + d which is 2, 4, 6 or 6, 4, 2<\/p>\n\n\n\n

5. If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

Sum of first three terms is 24<\/p>\n\n\n\n

Let us assume the first three terms are a \u2013 d, a, a + d [where a is the first term and d is the common difference]<\/p>\n\n\n\n

So, sum of first three terms is a \u2013 d + a + a + d = 24<\/p>\n\n\n\n

3a = 24<\/p>\n\n\n\n

a = 8<\/p>\n\n\n\n

It is given that the product of three terms is 440<\/p>\n\n\n\n

So a3<\/sup> \u2013 ad2<\/sup> = 440<\/p>\n\n\n\n

Substitute the value of a = 8, we get<\/p>\n\n\n\n

83<\/sup> \u2013 8d2<\/sup> = 440<\/p>\n\n\n\n

512 \u2013 8d2<\/sup> = 440<\/p>\n\n\n\n

72 = 8d2<\/sup><\/p>\n\n\n\n

d = 3 or d = \u2013 3<\/p>\n\n\n\n

Hence, the given terms are a \u2013 d, a, a + d which is 5, 8, 11<\/p>\n\n\n\n

6. The angles of a quadrilateral are in A.P. whose common difference is 10. Find the angles<\/strong><\/p>\n\n\n\n

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

Given: d = 10<\/p>\n\n\n\n

We know that the sum of all angles in a quadrilateral is 360<\/p>\n\n\n\n

Let us assume the angles are a \u2013 3d, a \u2013 d, a + d, a + 3d<\/p>\n\n\n\n

So, a \u2013 2d + a \u2013 d + a + d + a + 2d = 360<\/p>\n\n\n\n

4a = 360<\/p>\n\n\n\n

a = 90\u2026 (i)<\/p>\n\n\n\n

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

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

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

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

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

Hence, the angles are a \u2013 3d, a \u2013 d, a + d, a + 3d which is 75o<\/sup>, 85o<\/sup>, 95o<\/sup>, 105o<\/sup><\/p>\n\n\n\n

\n\n\n\n

EXERCISE 19.4 PAGE NO: 19.30<\/p>\n\n\n\n

1.<\/strong> Find the sum of the following arithmetic progressions:
(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-b, a-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>, \u2026 to n terms<\/strong><\/p>\n\n\n\n

(vii) (x \u2013 y)\/(x + y), (3x \u2013 2y)\/(x + y), (5x \u2013 3y)\/(x + y), \u2026 to n terms<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>50, 46, 42, \u2026. to 10 terms<\/p>\n\n\n\n

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

First term, a = a1<\/sub> = 50<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= 46 \u2013 50 = -4<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

Substitute the values of \u2018a\u2019 and \u2018d\u2019, we get<\/p>\n\n\n\n

S = 10\/2 (100 + (9) (-4))<\/p>\n\n\n\n

= 5 (100 \u2013 36)<\/p>\n\n\n\n

= 5 (64)<\/p>\n\n\n\n

= 320<\/p>\n\n\n\n

\u2234 The sum of the given AP is 320.<\/p>\n\n\n\n

(ii) <\/strong>1, 3, 5, 7, \u2026 to 12 terms<\/p>\n\n\n\n

n = 12<\/p>\n\n\n\n

First term, a = a1<\/sub> = 1<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= 3 \u2013 1 = 2<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

Substitute the values of \u2018a\u2019 and \u2018d\u2019, we get<\/p>\n\n\n\n

S = 12\/2 (2(1) + (12-1) (2))<\/p>\n\n\n\n

= 6 (2 + (11) (2))<\/p>\n\n\n\n

= 6 (2 + 22)<\/p>\n\n\n\n

= 6 (24)<\/p>\n\n\n\n

= 144<\/p>\n\n\n\n

\u2234 The sum of the given AP is 144.<\/p>\n\n\n\n

(iii) <\/strong>3, 9\/2, 6, 15\/2, \u2026 to 25 terms<\/p>\n\n\n\n

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

First term, a = a1<\/sub> = 3<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= 9\/2 \u2013 3 = (9 \u2013 6)\/2 = 3\/2<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

Substitute the values of \u2018a\u2019 and \u2018d\u2019, we get<\/p>\n\n\n\n

S = 25\/2 (2(3) + (25-1) (3\/2))<\/p>\n\n\n\n

= 25\/2 (6 + (24) (3\/2))<\/p>\n\n\n\n

= 25\/2 (6 + 36)<\/p>\n\n\n\n

= 25\/2 (42)<\/p>\n\n\n\n

= 25 (21)<\/p>\n\n\n\n

= 525<\/p>\n\n\n\n

\u2234 The sum of the given AP is 525.<\/p>\n\n\n\n

(iv) <\/strong>41, 36, 31, \u2026 to 12 terms<\/p>\n\n\n\n

n = 12<\/p>\n\n\n\n

First term, a = a1<\/sub> = 41<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= 36 \u2013 41 = -5<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

Substitute the values of \u2018a\u2019 and \u2018d\u2019, we get<\/p>\n\n\n\n

S = 12\/2 (2(41) + (12-1) (-5))<\/p>\n\n\n\n

= 6 (82 + (11) (-5))<\/p>\n\n\n\n

= 6 (82 \u2013 55)<\/p>\n\n\n\n

= 6 (27)<\/p>\n\n\n\n

= 162<\/p>\n\n\n\n

\u2234 The sum of the given AP is 162.<\/p>\n\n\n\n

(v) <\/strong>a+b, a-b, a-3b, \u2026 to 22 terms<\/p>\n\n\n\n

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

First term, a = a1<\/sub> = a+b<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= (a-b) \u2013 (a+b) = a-b-a-b = -2b<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

Substitute the values of \u2018a\u2019 and \u2018d\u2019, we get<\/p>\n\n\n\n

S = 22\/2 (2(a+b) + (22-1) (-2b))<\/p>\n\n\n\n

= 11 (2a + 2b + (21) (-2b))<\/p>\n\n\n\n

= 11 (2a + 2b \u2013 42b)<\/p>\n\n\n\n

= 11 (2a \u2013 40b)<\/p>\n\n\n\n

= 22a \u2013 440b<\/p>\n\n\n\n

\u2234 The sum of the given AP is 22a \u2013 440b.<\/p>\n\n\n\n

(vi) <\/strong>(x \u2013 y)2<\/sup>, (x2<\/sup> + y2<\/sup>), (x + y)2<\/sup>, \u2026 to n terms<\/p>\n\n\n\n

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

First term, a = a1<\/sub> = (x-y)2<\/sup><\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= (x2<\/sup> + y2<\/sup>) \u2013 (x-y)2<\/sup> = 2xy<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

Substitute the values of \u2018a\u2019 and \u2018d\u2019, we get<\/p>\n\n\n\n

S = n\/2 (2(x-y)2<\/sup> + (n-1) (2xy))<\/p>\n\n\n\n

= n\/2 (2 (x2<\/sup> + y2<\/sup> \u2013 2xy) + 2xyn \u2013 2xy)<\/p>\n\n\n\n

= n\/2 \u00d7 2 ((x2<\/sup> + y2<\/sup> \u2013 2xy) + xyn \u2013 xy)<\/p>\n\n\n\n

= n (x2<\/sup> + y2<\/sup> \u2013 3xy + xyn)<\/p>\n\n\n\n

\u2234 The sum of the given AP is n (x2<\/sup> + y2<\/sup> \u2013 3xy + xyn).<\/p>\n\n\n\n

(vii) <\/strong>(x \u2013 y)\/(x + y), (3x \u2013 2y)\/(x + y), (5x \u2013 3y)\/(x + y), \u2026 to n terms<\/p>\n\n\n\n

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

First term, a = a1<\/sub> = (x-y)\/(x+y)<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= (3x \u2013 2y)\/(x + y) \u2013 (x-y)\/(x+y) = (2x \u2013 y)\/(x+y)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

Substitute the values of \u2018a\u2019 and \u2018d\u2019, we get<\/p>\n\n\n\n

S = n\/2 (2((x-y)\/(x+y)) + (n-1) ((2x \u2013 y)\/(x+y)))<\/p>\n\n\n\n

= n\/2(x+y) {n (2x-y) \u2013 y}<\/p>\n\n\n\n

\u2234 The sum of the given AP is n\/2(x+y) {n (2x-y) \u2013 y}<\/p>\n\n\n\n

2.<\/strong> Find the sum of the following series:
(i) 2 + 5 + 8 + \u2026 + 182<\/strong><\/p>\n\n\n\n

(ii) 101 + 99 + 97 + \u2026 + 47<\/strong><\/p>\n\n\n\n

(iii) (a \u2013 b)2<\/sup> + (a2<\/sup> + b2<\/sup>) + (a + b)2<\/sup> + s\u2026. + [(a + b)2<\/sup> + 6ab]<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>2 + 5 + 8 + \u2026 + 182<\/p>\n\n\n\n

First term, a = a1<\/sub> = 2<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= 5 \u2013 2 = 3<\/p>\n\n\n\n

an<\/sub> term of given AP is 182<\/p>\n\n\n\n

an<\/sub> = a + (n-1) d<\/p>\n\n\n\n

182 = 2 + (n-1) 3<\/p>\n\n\n\n

182 = 2 + 3n \u2013 3<\/p>\n\n\n\n

182 = 3n \u2013 1<\/p>\n\n\n\n

3n = 182 + 1<\/p>\n\n\n\n

n = 183\/3<\/p>\n\n\n\n

= 61<\/p>\n\n\n\n

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

By using the formula,<\/p>\n\n\n\n

S = n\/2 (a + l)<\/p>\n\n\n\n

= 61\/2 (2 + 182)<\/p>\n\n\n\n

= 61\/2 (184)<\/p>\n\n\n\n

= 61 (92)<\/p>\n\n\n\n

= 5612<\/p>\n\n\n\n

\u2234 The sum of the series is 5612<\/p>\n\n\n\n

(ii) <\/strong>101 + 99 + 97 + \u2026 + 47<\/p>\n\n\n\n

First term, a = a1<\/sub> = 101<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= 99 \u2013 101 = -2<\/p>\n\n\n\n

an<\/sub> term of given AP is 47<\/p>\n\n\n\n

an<\/sub> = a + (n-1) d<\/p>\n\n\n\n

47 = 101 + (n-1)(-2)<\/p>\n\n\n\n

47 = 101 \u2013 2n + 2<\/p>\n\n\n\n

2n = 103 \u2013 47<\/p>\n\n\n\n

2n = 56<\/p>\n\n\n\n

n = 56\/2 = 28<\/p>\n\n\n\n

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

S = n\/2 (a + l)<\/p>\n\n\n\n

= 28\/2 (101 + 47)<\/p>\n\n\n\n

= 28\/2 (148)<\/p>\n\n\n\n

= 14 (148)<\/p>\n\n\n\n

= 2072<\/p>\n\n\n\n

\u2234 The sum of the series is 2072<\/p>\n\n\n\n

(iii) <\/strong>(a \u2013 b)2<\/sup> + (a2<\/sup> + b2<\/sup>) + (a + b)2<\/sup> + s\u2026. + [(a + b)2<\/sup> + 6ab]<\/p>\n\n\n\n

First term, a = a1<\/sub> = (a-b)2<\/sup><\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= (a2<\/sup> + b2<\/sup>) \u2013 (a \u2013 b)2<\/sup> = 2ab<\/p>\n\n\n\n

an<\/sub> term of given AP is [(a + b)2<\/sup> + 6ab]<\/p>\n\n\n\n

an<\/sub> = a + (n-1) d[(a + b)2<\/sup> + 6ab] = (a-b)2<\/sup> + (n-1)2ab<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 2ab + 6ab = a2<\/sup> + b2<\/sup> \u2013 2ab + 2abn \u2013 2ab<\/p>\n\n\n\n

a2<\/sup> + b2<\/sup> + 8ab \u2013 a2<\/sup> \u2013 b2<\/sup> + 2ab + 2ab = 2abn<\/p>\n\n\n\n

12ab = 2abn<\/p>\n\n\n\n

n = 12ab \/ 2ab<\/p>\n\n\n\n

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

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

S = n\/2 (a + l)<\/p>\n\n\n\n

= 6\/2 ((a-b)2<\/sup> + [(a + b)2<\/sup> + 6ab])<\/p>\n\n\n\n

= 3 (a2<\/sup> + b2<\/sup> \u2013 2ab + a2<\/sup> + b2<\/sup> + 2ab + 6ab)<\/p>\n\n\n\n

= 3 (2a2<\/sup> + 2b2<\/sup> + 6ab)<\/p>\n\n\n\n

= 3 \u00d7 2 (a2<\/sup> + b2<\/sup> + 3ab)<\/p>\n\n\n\n

= 6 (a2<\/sup> + b2<\/sup> + 3ab)<\/p>\n\n\n\n

\u2234 The sum of the series is 6 (a2<\/sup> + b2<\/sup> + 3ab)<\/p>\n\n\n\n

3. Find the sum of first n natural numbers.<\/strong><\/p>\n\n\n\n

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

Let AP be 1, 2, 3, 4, \u2026, n<\/p>\n\n\n\n

Here,<\/p>\n\n\n\n

First term, a = a1<\/sub> = 1<\/p>\n\n\n\n

Common difference, d = a2<\/sub> \u2013 a1 <\/sub>= 2 \u2013 1 = 1<\/p>\n\n\n\n

l = n<\/p>\n\n\n\n

So, the sum of n terms = S = n\/2 [2a + (n-1) d]<\/p>\n\n\n\n

= n\/2 [2(1) + (n-1) 1]<\/p>\n\n\n\n

= n\/2 [2 + n \u2013 1]<\/p>\n\n\n\n

= n\/2 [n \u2013 1]<\/p>\n\n\n\n

\u2234 The sum of the first n natural numbers is n(n-1)\/2<\/p>\n\n\n\n

4. Find the sum of all \u2013 natural numbers between 1 and 100, which are divisible by 2 or 5<\/strong><\/p>\n\n\n\n

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

The natural numbers which are divisible by 2 or 5 are:<\/p>\n\n\n\n

2 + 4 + 5 + 6 + 8 + 10 + \u2026 + 100 = (2 + 4 + 6 +\u2026+ 100) + (5 + 15 + 25 +\u2026+95)<\/p>\n\n\n\n

Now, (2 + 4 + 6 +\u2026+ 100) + (5 + 15 + 25 +\u2026+95) are AP with common difference of 2 and 10.<\/p>\n\n\n\n

So, for the 1st<\/sup> sequence => (2 + 4 + 6 +\u2026+ 100)<\/p>\n\n\n\n

a = 2, d = 4-2 = 2, an<\/sub> = 100<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

100 = 2 + (n-1)2<\/p>\n\n\n\n

100 = 2 + 2n \u2013 2<\/p>\n\n\n\n

2n = 100<\/p>\n\n\n\n

n = 100\/2<\/p>\n\n\n\n

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

So now, S = n\/2 (2a + (n-1)d)<\/p>\n\n\n\n

= 50\/2 (2(2) + (50-1)2)<\/p>\n\n\n\n

= 25 (4 + 49(2))<\/p>\n\n\n\n

= 25 (4 + 98)<\/p>\n\n\n\n

= 2550<\/p>\n\n\n\n

Again, for the 2nd<\/sup> sequence, (5 + 15 + 25 +\u2026+95)<\/p>\n\n\n\n

a = 5, d = 15-5 = 10, an<\/sub> = 95<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

95 = 5 + (n-1)10<\/p>\n\n\n\n

95 = 5 + 10n \u2013 10<\/p>\n\n\n\n

10n = 95 +10 \u2013 5<\/p>\n\n\n\n

10n = 100<\/p>\n\n\n\n

n = 100\/10<\/p>\n\n\n\n

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

So now, S = n\/2 (2a + (n-1)d)<\/p>\n\n\n\n

= 10\/2 (2(5) + (10-1)10)<\/p>\n\n\n\n

= 5 (10 + 9(10))<\/p>\n\n\n\n

= 5 (10 + 90)<\/p>\n\n\n\n

= 500<\/p>\n\n\n\n

\u2234 The sum of the numbers divisible by 2 or 5 is: 2550 + 500 = 3050<\/p>\n\n\n\n

5. Find the sum of first n odd natural numbers.<\/strong><\/p>\n\n\n\n

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

Given an AP of first n odd natural numbers whose first term a is 1, and common difference d is 3<\/p>\n\n\n\n

The sequence is 1, 3, 5, 7\u2026\u2026n<\/p>\n\n\n\n

a = 1, d = 3-1 = 2, n = n<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

S = n\/2 [2a + (n-1)d]<\/p>\n\n\n\n

= n\/2 [2(1) + (n-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

\u2234 The sum of the first n odd natural numbers is n2<\/sup>.<\/p>\n\n\n\n

6. Find the sum of all odd numbers between 100 and 200<\/strong><\/p>\n\n\n\n

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

The series is 101, 103, 105, \u2026, 199<\/p>\n\n\n\n

Let the number of terms be n<\/p>\n\n\n\n

So, a = 101, d = 103 \u2013 101 = 2, an<\/sub> = 199<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

199 = 101 + (n-1)2<\/p>\n\n\n\n

199 = 101 + 2n \u2013 2<\/p>\n\n\n\n

2n = 199 \u2013 101 + 2<\/p>\n\n\n\n

2n = 100<\/p>\n\n\n\n

n = 100\/2<\/p>\n\n\n\n

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

By using the formula,<\/p>\n\n\n\n

The sum of n terms = S = n\/2[a + l]<\/p>\n\n\n\n

= 50\/2 [101 + 199]<\/p>\n\n\n\n

= 25 [300]<\/p>\n\n\n\n

= 7500<\/p>\n\n\n\n

\u2234 The sum of the odd numbers between 100 and 200 is 7500.<\/p>\n\n\n\n

7. Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667<\/strong><\/p>\n\n\n\n

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

The odd numbers between 1 and 1000 divisible by 3 are 3, 9, 15,\u2026,999<\/p>\n\n\n\n

Let the number of terms be \u2018n\u2019, so the nth term is 999<\/p>\n\n\n\n

a = 3, d = 9-3 = 6, an<\/sub> = 999<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

999 = 3 + (n-1)6<\/p>\n\n\n\n

999 = 3 + 6n \u2013 6<\/p>\n\n\n\n

6n = 999 + 6 \u2013 3<\/p>\n\n\n\n

6n = 1002<\/p>\n\n\n\n

n = 1002\/6<\/p>\n\n\n\n

= 167<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of n terms, S = n\/2 [a + l]<\/p>\n\n\n\n

= 167\/2 [3 + 999]<\/p>\n\n\n\n

= 167\/2 [1002]<\/p>\n\n\n\n

= 167 [501]<\/p>\n\n\n\n

= 83667<\/p>\n\n\n\n

\u2234 The sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

8. Find the sum of all integers between 84 and 719, which are multiples of 5<\/strong><\/p>\n\n\n\n

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

The series is 85, 90, 95, \u2026, 715<\/p>\n\n\n\n

Let there be \u2018n\u2019 terms in the AP<\/p>\n\n\n\n

So, a = 85, d = 90-85 = 5, an<\/sub> = 715<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

715 = 85 + (n-1)5<\/p>\n\n\n\n

715 = 85 + 5n \u2013 5<\/p>\n\n\n\n

5n = 715 \u2013 85 + 5<\/p>\n\n\n\n

5n = 635<\/p>\n\n\n\n

n = 635\/5<\/p>\n\n\n\n

= 127<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of n terms, S = n\/2 [a + l]<\/p>\n\n\n\n

= 127\/2 [85 + 715]<\/p>\n\n\n\n

= 127\/2 [800]<\/p>\n\n\n\n

= 127 [400]<\/p>\n\n\n\n

= 50800<\/p>\n\n\n\n

\u2234 The sum of all integers between 84 and 719, which are multiples of 5 is 50800.<\/p>\n\n\n\n

9. Find the sum of all integers between 50 and 500 which are divisible by 7<\/strong><\/p>\n\n\n\n

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

The series of integers divisible by 7 between 50 and 500 are 56, 63, 70, \u2026, 497<\/p>\n\n\n\n

Let the number of terms be \u2018n\u2019<\/p>\n\n\n\n

So, a = 56, d = 63-56 = 7, an<\/sub> = 497<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

497 = 56 + (n-1)7<\/p>\n\n\n\n

497 = 56 + 7n \u2013 7<\/p>\n\n\n\n

7n = 497 \u2013 56 + 7<\/p>\n\n\n\n

7n = 448<\/p>\n\n\n\n

n = 448\/7<\/p>\n\n\n\n

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

By using the formula,<\/p>\n\n\n\n

Sum of n terms, S = n\/2 [a + l]<\/p>\n\n\n\n

= 64\/2 [56 + 497]<\/p>\n\n\n\n

= 32 [553]<\/p>\n\n\n\n

= 17696<\/p>\n\n\n\n

\u2234 The sum of all integers between 50 and 500 which are divisible by 7 is 17696.<\/p>\n\n\n\n

10. Find the sum of all even integers between 101 and 999<\/strong><\/p>\n\n\n\n

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

We know that all even integers will have a common difference of 2.<\/p>\n\n\n\n

So, AP is 102, 104, 106, \u2026, 998<\/p>\n\n\n\n

We know, a = 102, d = 104 \u2013 102 = 2, an<\/sub> = 998<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

998 = 102 + (n-1)2<\/p>\n\n\n\n

998 = 102 + 2n \u2013 2<\/p>\n\n\n\n

2n = 998 \u2013 102 + 2<\/p>\n\n\n\n

2n = 898<\/p>\n\n\n\n

n = 898\/2<\/p>\n\n\n\n

= 449<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sum of n terms, S = n\/2 [a + l]<\/p>\n\n\n\n

= 449\/2 [102 + 998]<\/p>\n\n\n\n

= 449\/2 [1100]<\/p>\n\n\n\n

= 449 [550]<\/p>\n\n\n\n

= 246950<\/p>\n\n\n\n

\u2234 The sum of all even integers between 101 and 999 is 246950.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 19.5 PAGE NO: 19.42<\/p>\n\n\n\n

1. If 1\/a, 1\/b, 1\/c are in A.P., prove that:<\/strong><\/p>\n\n\n\n

(i) (b+c)\/a, (c+a)\/b, (a+b)\/c are in A.P.<\/strong><\/p>\n\n\n\n

(ii) a(b + c), b(c + a), c(a + b) are in A.P.<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>(b+c)\/a, (c+a)\/b, (a+b)\/c are in A.P.<\/p>\n\n\n\n

We know that, if a, b, c are in AP, then b \u2013 a = c \u2013 b<\/p>\n\n\n\n

If, 1\/a, 1\/b, 1\/c are in AP<\/p>\n\n\n\n

Then, 1\/b \u2013 1\/a = 1\/c \u2013 1\/b<\/p>\n\n\n\n

If (b+c)\/a, (c+a)\/b, (a+b)\/c are in AP<\/p>\n\n\n\n

Then, (c+a)\/b \u2013 (b+c)\/a = (a+b)\/c \u2013 (c+a)\/b<\/p>\n\n\n\n

Let us take LCM<\/p>\n\n\n\n

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

Since, 1\/a, 1\/b, 1\/c are in AP<\/p>\n\n\n\n

1\/b \u2013 1\/a = 1\/c \u2013 1\/b<\/p>\n\n\n\n

C (b \u2013 a) = a (b-c)<\/p>\n\n\n\n

Hence, the given terms are in AP.<\/p>\n\n\n\n

(ii) <\/strong>a(b + c), b(c + a), c(a + b) are in A.P.<\/p>\n\n\n\n

We know that if, b(c + a) \u2013 a(b+c) = c(a+b) \u2013 b(c+a)<\/p>\n\n\n\n

Consider LHS:<\/p>\n\n\n\n

b(c + a) \u2013 a(b+c)<\/p>\n\n\n\n

Upon simplification we get,<\/p>\n\n\n\n

b(c + a) \u2013 a(b+c) = bc + ba \u2013 ab \u2013 ac<\/p>\n\n\n\n

= c (b-a)<\/p>\n\n\n\n

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

c(a+b) \u2013 b(c+a) = ca + cb \u2013 bc \u2013 ba<\/p>\n\n\n\n

= a (c-b)<\/p>\n\n\n\n

We know,<\/p>\n\n\n\n

1\/a, 1\/b, 1\/c are in AP<\/p>\n\n\n\n

So, 1\/a \u2013 1\/b = 1\/b \u2013 1\/c<\/p>\n\n\n\n

Or c(b-a) = a(c-b)<\/p>\n\n\n\n

Hence, given terms are in AP.<\/p>\n\n\n\n

2. If a2<\/sup>, b2<\/sup>, c2<\/sup> are in AP., prove that a\/(b+c), b\/(c+a), c\/(a+b) are in AP.<\/strong><\/p>\n\n\n\n

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

If a2<\/sup>, b2<\/sup>, c2<\/sup> are in AP then, b2<\/sup> \u2013 a2<\/sup> = c2<\/sup> \u2013 b2<\/sup><\/p>\n\n\n\n

If a\/(b+c), b\/(c+a), c\/(a+b) are in AP then,<\/p>\n\n\n\n

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

Let us take LCM on both the sides we get,<\/p>\n\n\n\n

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

Since, b2<\/sup> \u2013 a2<\/sup> = c2<\/sup> \u2013 b2<\/sup><\/p>\n\n\n\n

Substituting b2<\/sup> \u2013 a2<\/sup>= c2<\/sup> \u2013 b2<\/sup> in above, we get<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

Hence, given terms are in AP<\/p>\n\n\n\n

3. If a, b, c are in A.P., then show that:
(i) a2<\/sup>(b + c), b2<\/sup>(c + a), c2<\/sup>(a + b) are also in A.P.
(ii) b + c \u2013 a, c + a \u2013 b, a + b \u2013 c are in A.P.
(iii) bc \u2013 a2<\/sup>, ca \u2013 b2<\/sup>, ab \u2013 c2<\/sup> are in A.P.<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>a2<\/sup>(b + c), b2<\/sup>(c + a), c2<\/sup>(a + b) are also in A.P.<\/p>\n\n\n\n

If b2<\/sup>(c + a) \u2013 a2<\/sup>(b + c) = c2<\/sup>(a + b) \u2013 b2<\/sup>(c + a)<\/p>\n\n\n\n

b2<\/sup>c + b2<\/sup>a \u2013 a2<\/sup>b \u2013 a2<\/sup>c = c2<\/sup>a + c2<\/sup>b \u2013 b2<\/sup>a \u2013 b2<\/sup>c<\/p>\n\n\n\n

Given, b \u2013 a = c \u2013 b<\/p>\n\n\n\n

And since a, b, c are in AP,<\/p>\n\n\n\n

c(b2<\/sup> \u2013 a2<\/sup> ) + ab(b \u2013 a) = a(c2<\/sup> \u2013 b2<\/sup> ) + bc(c \u2013 b)<\/p>\n\n\n\n

(b \u2013 a) (ab + bc + ca) = (c \u2013 b) (ab + bc + ca)<\/p>\n\n\n\n

Upon cancelling, ab + bc + ca from both sides<\/p>\n\n\n\n

b \u2013 a = c \u2013 b<\/p>\n\n\n\n

2b = c + a [which is true]<\/p>\n\n\n\n

Hence, given terms are in AP<\/p>\n\n\n\n

(ii)<\/strong> b + c \u2013 a, c + a \u2013 b, a + b \u2013 c are in A.P.<\/p>\n\n\n\n

If (c + a \u2013 b) \u2013 (b + c \u2013 a) = (a + b \u2013 c) \u2013 (c + a \u2013 b)<\/p>\n\n\n\n

Then, b + c \u2013 a, c + a \u2013 b, a + b \u2013 c are in A.P.<\/p>\n\n\n\n

Let us consider LHS and RHS<\/p>\n\n\n\n

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

2a \u2013 2b = 2b \u2013 2c<\/p>\n\n\n\n

b \u2013 a = c \u2013 b<\/p>\n\n\n\n

And since a, b, c are in AP,<\/p>\n\n\n\n

b \u2013 a = c \u2013 b<\/p>\n\n\n\n

Hence, given terms are in AP.<\/p>\n\n\n\n

(iii)<\/strong> bc \u2013 a2<\/sup>, ca \u2013 b2<\/sup>, ab \u2013 c2<\/sup> are in A.P.<\/p>\n\n\n\n

If (ca \u2013 b2<\/sup>) \u2013 (bc \u2013 a2<\/sup>) = (ab \u2013 c2<\/sup>) \u2013 (ca \u2013 b2<\/sup>)<\/p>\n\n\n\n

Then, bc \u2013 a2<\/sup>, ca \u2013 b2<\/sup>, ab \u2013 c2<\/sup> are in A.P.<\/p>\n\n\n\n

Let us consider LHS and RHS<\/p>\n\n\n\n

(ca \u2013 b2<\/sup>) \u2013 (bc \u2013 a2<\/sup>) = (ab \u2013 c2<\/sup>) \u2013 (ca \u2013 b2<\/sup>)<\/p>\n\n\n\n

(a \u2013 b2<\/sup> \u2013 bc + a2<\/sup>) = (ab \u2013 c2<\/sup> \u2013 ca + b2<\/sup>)<\/p>\n\n\n\n

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

a \u2013 b = b \u2013 c<\/p>\n\n\n\n

And since a, b, c are in AP,<\/p>\n\n\n\n

b \u2013 c = a \u2013 b<\/p>\n\n\n\n

Hence, given terms are in AP<\/p>\n\n\n\n

4. If (b+c)\/a, (c+a)\/b, (a+b)\/c are in AP., prove that:<\/strong><\/p>\n\n\n\n

(i) 1\/a, 1\/b, 1\/c are in AP<\/strong><\/p>\n\n\n\n

(ii) bc, ca, ab are in AP<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>1\/a, 1\/b, 1\/c are in AP<\/p>\n\n\n\n

If 1\/a, 1\/b, 1\/c are in AP then,<\/p>\n\n\n\n

1\/b \u2013 1\/a = 1\/c \u2013 1\/b<\/p>\n\n\n\n

Let us consider LHS:<\/p>\n\n\n\n

1\/b \u2013 1\/a = (a-b)\/ab<\/p>\n\n\n\n

= c(a-b)\/abc [by multiplying with \u2018c\u2019 on both the numerator and denominator]<\/p>\n\n\n\n

Let us consider RHS:<\/p>\n\n\n\n

1\/c \u2013 1\/b = (b-c)\/bc<\/p>\n\n\n\n

= a(b-c)\/bc [by multiplying with \u2018a\u2019 on both the numerator and denominator]<\/p>\n\n\n\n

Since, (b+c)\/a, (c+a)\/b, (a+b)\/c are in AP<\/p>\n\n\n\n

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

(ii) <\/strong>bc, ca, ab are in AP<\/p>\n\n\n\n

If bc, ca, ab are in AP then,<\/p>\n\n\n\n

ca \u2013 bc = ab \u2013 ca<\/p>\n\n\n\n

c (a-b) = a (b-c)<\/p>\n\n\n\n

If 1\/a, 1\/b, 1\/c are in AP then,<\/p>\n\n\n\n

1\/b \u2013 1\/a = 1\/c \u2013 1\/b<\/p>\n\n\n\n

c (a-b) = a (b-c)<\/p>\n\n\n\n

Hence, the given terms are in AP<\/p>\n\n\n\n

5. If a, b, c are in A.P., prove that:
(i) (a \u2013 c)2<\/sup> = 4 (a \u2013 b) (b \u2013 c)<\/strong><\/p>\n\n\n\n

(ii) a2<\/sup> + c2<\/sup> + 4ac = 2 (ab + bc + ca)<\/strong><\/p>\n\n\n\n

(iii) a3<\/sup> + c3<\/sup> + 6abc = 8b3<\/sup><\/strong><\/p>\n\n\n\n

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

(i) <\/strong>(a \u2013 c)2<\/sup> = 4 (a \u2013 b) (b \u2013 c)<\/p>\n\n\n\n

Let us expand the above expression<\/p>\n\n\n\n

a2<\/sup> + c2<\/sup> \u2013 2ac = 4(ab \u2013 ac \u2013 b2<\/sup> + bc)<\/p>\n\n\n\n

a2<\/sup> + 4c2<\/sup>b2<\/sup> + 2ac \u2013 4ab \u2013 4bc = 0<\/p>\n\n\n\n

(a + c \u2013 2b)2<\/sup> = 0<\/p>\n\n\n\n

a + c \u2013 2b = 0<\/p>\n\n\n\n

Since a, b, c are in AP<\/p>\n\n\n\n

b \u2013 a = c \u2013 b<\/p>\n\n\n\n

a + c \u2013 2b = 0<\/p>\n\n\n\n

a + c = 2b<\/p>\n\n\n\n

Hence, (a \u2013 c)2<\/sup> = 4 (a \u2013 b) (b \u2013 c)<\/p>\n\n\n\n

(ii) <\/strong>a2<\/sup> + c2<\/sup> + 4ac = 2 (ab + bc + ca)<\/p>\n\n\n\n

Let us expand the above expression<\/p>\n\n\n\n

a2<\/sup> + c2<\/sup> + 4ac = 2 (ab + bc + ca)<\/p>\n\n\n\n

a2<\/sup> + c2<\/sup> + 2ac \u2013 2ab \u2013 2bc = 0<\/p>\n\n\n\n

(a + c \u2013 b)2<\/sup> \u2013 b2<\/sup> = 0<\/p>\n\n\n\n

a + c \u2013 b = b<\/p>\n\n\n\n

a + c \u2013 2b = 0<\/p>\n\n\n\n

2b = a+c<\/p>\n\n\n\n

b = (a+c)\/2<\/p>\n\n\n\n

Since a, b, c are in AP<\/p>\n\n\n\n

b \u2013 a = c \u2013 b<\/p>\n\n\n\n

b = (a+c)\/2<\/p>\n\n\n\n

Hence, a2<\/sup> + c2<\/sup> + 4ac = 2 (ab + bc + ca)<\/p>\n\n\n\n

(iii) <\/strong>a3<\/sup> + c3<\/sup> + 6abc = 8b3<\/sup><\/p>\n\n\n\n

Let us expand the above expression<\/p>\n\n\n\n

a3<\/sup> + c3<\/sup> + 6abc = 8b3<\/sup><\/p>\n\n\n\n

a3<\/sup> + c3<\/sup> \u2013 (2b)3<\/sup> + 6abc = 0<\/p>\n\n\n\n

a3<\/sup> + (-2b)3<\/sup> + c3<\/sup> + 3a(-2b)c = 0<\/p>\n\n\n\n

Since, if a + b + c = 0, a3<\/sup> + b3<\/sup> + c3<\/sup> = 3abc<\/p>\n\n\n\n

(a \u2013 2b + c)3<\/sup> = 0<\/p>\n\n\n\n

a \u2013 2b + c = 0<\/p>\n\n\n\n

a + c = 2b<\/p>\n\n\n\n

b = (a+c)\/2<\/p>\n\n\n\n

Since a, b, c are in AP<\/p>\n\n\n\n

a \u2013 b = c \u2013 b<\/p>\n\n\n\n

b = (a+c)\/2<\/p>\n\n\n\n

Hence, a3<\/sup> + c3<\/sup> + 6abc = 8b3<\/sup><\/p>\n\n\n\n

6. If a(1\/b + 1\/c), b(1\/c + 1\/a), c(1\/a + 1\/b) are in AP., prove that a, b, c are in AP.<\/strong><\/p>\n\n\n\n

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

Here, we know a(1\/b + 1\/c), b(1\/c + 1\/a), c(1\/a + 1\/b) are in AP<\/p>\n\n\n\n

Also, a(1\/b + 1\/c) + 1, b(1\/c + 1\/a) + 1, c(1\/a + 1\/b) + 1 are in AP<\/p>\n\n\n\n

Let us take LCM for each expression then we get,<\/p>\n\n\n\n

(ac+ab+bc)\/bc , (ab+bc+ac)\/ac, (cb+ac+ab)\/ab are in AP<\/p>\n\n\n\n

1\/bc, 1\/ac, 1\/ab are in AP<\/p>\n\n\n\n

Let us multiply numerator with \u2018abc\u2019, we get<\/p>\n\n\n\n

abc\/bc, abc\/ac, abc\/ab are in AP<\/p>\n\n\n\n

\u2234 a, b, c are in AP.<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

7. Show that x2<\/sup> + xy + y2<\/sup>, z2<\/sup> + zx + x2<\/sup> and y2<\/sup> + yz + z2<\/sup> are in consecutive terms of an A.P., if x, y and z are in A.P.<\/strong><\/p>\n\n\n\n

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

x, y, z are in AP<\/p>\n\n\n\n

Given, x2<\/sup> + xy + y2<\/sup>, z2<\/sup> + zx + x2<\/sup> and y2<\/sup> + yz + z2<\/sup> are in AP<\/p>\n\n\n\n

(z2<\/sup> + zx + x2<\/sup>) \u2013 (x2<\/sup> + xy + y2<\/sup>) = (y2<\/sup> + yz + z2<\/sup>) \u2013 (z2<\/sup> + zx + x2<\/sup>)<\/p>\n\n\n\n

Let d = common difference,<\/p>\n\n\n\n

So, Y = x + d and x = x + 2d<\/p>\n\n\n\n

Let us consider the LHS:<\/p>\n\n\n\n

(z2<\/sup> + zx + x2<\/sup>) \u2013 (x2<\/sup> + xy + y2<\/sup>)<\/p>\n\n\n\n

z2<\/sup> + zx \u2013 xy \u2013 y2<\/sup><\/p>\n\n\n\n

(x + 2d)2<\/sup> + (x + 2d)x \u2013 x(x + d) \u2013 (x + d)2<\/sup><\/p>\n\n\n\n

x2<\/sup> + 4xd + 4d2<\/sup> + x2<\/sup> + 2xd \u2013 x2<\/sup> \u2013 xd \u2013 x2<\/sup> \u2013 2xd \u2013 d2<\/sup><\/p>\n\n\n\n

3xd + 3d2<\/sup><\/p>\n\n\n\n

Now, let us consider RHS:<\/p>\n\n\n\n

(y2<\/sup> + yz + z2<\/sup>) \u2013 (z2<\/sup> + zx + x2<\/sup>)<\/p>\n\n\n\n

y2<\/sup> + yz \u2013 zx \u2013 x2<\/sup><\/p>\n\n\n\n

(x + d)2<\/sup> + (x + d)(x + 2d) \u2013 (x + 2d)x \u2013 x2<\/sup><\/p>\n\n\n\n

x2<\/sup> + 2dx + d2<\/sup> + x2<\/sup> + 2dx + xd + 2d2<\/sup> \u2013 x2<\/sup> \u2013 2dx \u2013 x2<\/sup><\/p>\n\n\n\n

3xd + 3d2<\/sup><\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

\u2234 x2<\/sup> + xy + y2<\/sup>, z2<\/sup> + zx + x2<\/sup> and y2<\/sup> + yz + z2<\/sup> are in consecutive terms of A.P<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 19.6 PAGE NO: 19.46<\/p>\n\n\n\n

1. Find the A.M. between:
(i) 7 and 13 (ii) 12 and \u2013 8 (iii) (x \u2013 y) and (x + y)<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> Let A be the Arithmetic mean<\/p>\n\n\n\n

Then 7, A, 13 are in AP<\/p>\n\n\n\n

Now, let us solve<\/p>\n\n\n\n

A-7 = 13-A<\/p>\n\n\n\n

2A = 13 + 7<\/p>\n\n\n\n

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

(ii)<\/strong> Let A be the Arithmetic mean<\/p>\n\n\n\n

Then 12, A, \u2013 8 are in AP<\/p>\n\n\n\n

Now, let us solve<\/p>\n\n\n\n

A \u2013 12 = \u2013 8 \u2013 A<\/p>\n\n\n\n

2A = 12 + 8<\/p>\n\n\n\n

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

(iii)<\/strong> Let A be the Arithmetic mean<\/p>\n\n\n\n

Then x \u2013 y, A, x + y are in AP<\/p>\n\n\n\n

Now, let us solve<\/p>\n\n\n\n

A \u2013 (x \u2013 y) = (x + y) \u2013 A<\/p>\n\n\n\n

2A = x + y + x \u2013 y<\/p>\n\n\n\n

A = x<\/p>\n\n\n\n

2. Insert 4 A.M.s between 4 and 19.<\/strong><\/p>\n\n\n\n

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

Let A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub> be the 4 AM Between 4 and 19<\/p>\n\n\n\n

Then, 4, A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, 19 are in AP.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

= (19 \u2013 4) \/ (4 + 1)<\/p>\n\n\n\n

= 15\/5<\/p>\n\n\n\n

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

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

A1<\/sub> = a + d = 4 + 3 = 7<\/p>\n\n\n\n

A2<\/sub> = A1 + d = 7 + 3 = 10<\/p>\n\n\n\n

A3<\/sub> = A2 + d = 10 + 3 = 13<\/p>\n\n\n\n

A4<\/sub> = A3 + d = 13 + 3 = 16<\/p>\n\n\n\n

3. Insert 7 A.M.s between 2 and 17.<\/strong><\/p>\n\n\n\n

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

Let A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub>, A6<\/sub>, A7<\/sub> be the 7 AMs between 2 and 17<\/p>\n\n\n\n

Then, 2, A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub>, A6<\/sub>, A7<\/sub>, 17 are in AP<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

an<\/sub> = 17, a = 2, n = 9<\/p>\n\n\n\n

so,<\/p>\n\n\n\n

17 = 2 + (9 \u2013 1)d<\/p>\n\n\n\n

17 = 2 + 9d \u2013 d<\/p>\n\n\n\n

17 = 2 + 8d<\/p>\n\n\n\n

8d = 17 \u2013 2<\/p>\n\n\n\n

8d = 15<\/p>\n\n\n\n

d = 15\/8<\/p>\n\n\n\n

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

A1<\/sub> = a + d = 2 + 15\/8 = 31\/8<\/p>\n\n\n\n

A2<\/sub> = A1<\/sub> + d = 31\/8 + 15\/8 = 46\/8<\/p>\n\n\n\n

A3<\/sub> = A2<\/sub> + d = 46\/8 + 15\/8 = 61\/8<\/p>\n\n\n\n

A4<\/sub> = A3<\/sub> + d = 61\/8 + 15\/8 = 76\/8<\/p>\n\n\n\n

A5<\/sub> = A4<\/sub> + d = 76\/8 + 15\/8 = 91\/8<\/p>\n\n\n\n

A6<\/sub> = A5<\/sub> + d = 91\/8 + 15\/8 = 106\/8<\/p>\n\n\n\n

A7<\/sub> = A6<\/sub> + d = 106\/8 + 15\/8 = 121\/8<\/p>\n\n\n\n

\u2234 the 7 AMs between 2 and 7 are 31\/8, 46\/8, 61\/8, 76\/8, 91\/8, 106\/8, 121\/8<\/p>\n\n\n\n

4. Insert six A.M.s between 15 and \u2013 13.<\/strong><\/p>\n\n\n\n

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

Let A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub>, A6<\/sub> be the 7 AM between 15 and \u2013 13<\/p>\n\n\n\n

Then, 15, A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub>, A6<\/sub>, \u2013 13 are in AP<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

an<\/sub> = -13, a = 15, n = 8<\/p>\n\n\n\n

so,<\/p>\n\n\n\n

-13 = 15 + (8 \u2013 1)d<\/p>\n\n\n\n

-13 = 15 + 7d<\/p>\n\n\n\n

7d = -13 \u2013 15<\/p>\n\n\n\n

7d = -28<\/p>\n\n\n\n

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

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

A1<\/sub> = a + d = 15 \u2013 4 = 11<\/p>\n\n\n\n

A2<\/sub> = A1 + d = 11 \u2013 4 = 7<\/p>\n\n\n\n

A3<\/sub> = A2 + d = 7 \u2013 4 = 3<\/p>\n\n\n\n

A4<\/sub> = A3 + d = 3 \u2013 4 = -1<\/p>\n\n\n\n

A5<\/sub> = A4 + d = -1 \u2013 4 = -5<\/p>\n\n\n\n

A6<\/sub> = A5 + d = -5 \u2013 4 = -9<\/p>\n\n\n\n

5. There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3: 1. Find the value of n.<\/strong><\/p>\n\n\n\n

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

Let the series be 3, A1<\/sub>, A2<\/sub>, A3<\/sub>, \u2026\u2026.., An<\/sub>, 17<\/p>\n\n\n\n

Given, an<\/sub>\/a1<\/sub> = 3\/1<\/p>\n\n\n\n

We know total terms in AP are n + 2<\/p>\n\n\n\n

So, 17 is the (n + 2)th term<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

An<\/sub> = 17, a = 3<\/p>\n\n\n\n

So, 17 = 3 + (n + 2 \u2013 1)d<\/p>\n\n\n\n

17 = 3 + (n + 1)d<\/p>\n\n\n\n

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

14 = (n + 1)d<\/p>\n\n\n\n

d = 14\/(n+1)<\/p>\n\n\n\n

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

An<\/sub> = 3 + 14\/(n+1) = (17n + 3) \/ (n+1)<\/p>\n\n\n\n

A1<\/sub> = 3 + d = (3n+17)\/(n+1)<\/p>\n\n\n\n

Since,<\/p>\n\n\n\n

an<\/sub>\/a1<\/sub> = 3\/1<\/p>\n\n\n\n

(17n + 3)\/ (3n+17) = 3\/1<\/p>\n\n\n\n

17n + 3 = 3(3n + 17)<\/p>\n\n\n\n

17n + 3 = 9n + 51<\/p>\n\n\n\n

17n \u2013 9n = 51 \u2013 3<\/p>\n\n\n\n

8n = 48<\/p>\n\n\n\n

n = 48\/8<\/p>\n\n\n\n

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

\u2234 There are 6 terms in the AP<\/p>\n\n\n\n

6. Insert A.M.s between 7 and 71 in such a way that the 5th<\/sup> A.M. is 27. Find the number of A.M.s.<\/strong><\/p>\n\n\n\n

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

Let the series be 7, A1<\/sub>, A2<\/sub>, A3<\/sub>, \u2026\u2026.., An<\/sub>, 71<\/p>\n\n\n\n

We know total terms in AP are n + 2<\/p>\n\n\n\n

So 71 is the (n + 2)th term<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

An<\/sub> = 71, n = 6<\/p>\n\n\n\n

A6<\/sub> = a + (6 \u2013 1)d<\/p>\n\n\n\n

a + 5d = 27 (5th<\/sup> term)<\/p>\n\n\n\n

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

so,<\/p>\n\n\n\n

71 = (n + 2)th term<\/p>\n\n\n\n

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

71 = 7 + n(4)<\/p>\n\n\n\n

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

\u2234 There are 15 terms in AP<\/p>\n\n\n\n

7. If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.<\/strong><\/p>\n\n\n\n

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

Let a and b be the first and last terms<\/p>\n\n\n\n

The series be a, A1<\/sub>, A2<\/sub>, A3<\/sub>, \u2026\u2026.., An<\/sub>, b<\/p>\n\n\n\n

We know, Mean = (a+b)\/2<\/p>\n\n\n\n

Mean of A1<\/sub> and An<\/sub> = (A1<\/sub> + An<\/sub>)\/2<\/p>\n\n\n\n

A1<\/sub> = a+d<\/p>\n\n\n\n

An<\/sub> = a \u2013 d<\/p>\n\n\n\n

So, AM = (a+d+b-d)\/2<\/p>\n\n\n\n

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

AM between A2<\/sub> and An-1<\/sub> = (a+2d+b-2d)\/2<\/p>\n\n\n\n

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

Similarly, (a + b)\/2 is constant for all such numbers<\/p>\n\n\n\n

Hence, AM = (a + b)\/2<\/p>\n\n\n\n

8. If x, y, z are in A.P. and A1<\/sub>is the A.M. of x and y, and A2<\/sub> is the A.M. of y and z, then prove that the A.M. of A1<\/sub> and A2<\/sub> is y.<\/strong><\/p>\n\n\n\n

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

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

A1<\/sub> = AM of x and y<\/p>\n\n\n\n

And A2<\/sub> = AM of y and z<\/p>\n\n\n\n

So, A1<\/sub> = (x+y)\/2<\/p>\n\n\n\n

A2<\/sub> = (y+x)\/2<\/p>\n\n\n\n

AM of A1<\/sub> and A2<\/sub> = (A1<\/sub> + A2<\/sub>)\/2<\/p>\n\n\n\n

= [(x+y)\/2 + (y+z)\/2]\/2<\/p>\n\n\n\n

= [x+y+y+z]\/2<\/p>\n\n\n\n

= [x+2y+z]\/2<\/p>\n\n\n\n

Since x, y, z are in AP, y = (x+z)\/2<\/p>\n\n\n\n

AM = [(x + z\/2) + (2y\/2)]\/2<\/p>\n\n\n\n

= (y + y)\/2<\/p>\n\n\n\n

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

= y<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

9. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P<\/strong><\/p>\n\n\n\n

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

Let A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub> be the 5 numbers between 8 and 26<\/p>\n\n\n\n

Then, 8, A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub>, 26 are in AP<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

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

An<\/sub> = 26, a = 8, n = 7<\/p>\n\n\n\n

26 = 8 + (7 \u2013 1)d<\/p>\n\n\n\n

26 = 8 + 6d<\/p>\n\n\n\n

6d = 26 \u2013 8<\/p>\n\n\n\n

6d = 18<\/p>\n\n\n\n

d = 18\/6<\/p>\n\n\n\n

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

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

A1<\/sub> = a + d = 8 + 3 = 11<\/p>\n\n\n\n

A2<\/sub> = A1<\/sub> + d = 11 + 3 = 14<\/p>\n\n\n\n

A3<\/sub> = A2<\/sub> + d = 14 + 3 = 17<\/p>\n\n\n\n

A4<\/sub> = A3<\/sub> + d = 17 + 3 = 20<\/p>\n\n\n\n

A5<\/sub> = A4<\/sub> + d = 20 + 3 = 23<\/p>\n\n\n\n

EXERCISE 19.7 PAGE NO: 19.48<\/p>\n\n\n\n

1. A man saved \u20b9 16500 in ten years. In each year after the first he saved \u20b9 100\/- more than he did in the preceding year. How much did he saved in the first year?<\/strong><\/p>\n\n\n\n

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

Given: A man saved \u20b916500 in ten years<\/p>\n\n\n\n

Let \u20b9 x be his savings in the first year<\/p>\n\n\n\n

His savings increased by \u20b9 100 every year.<\/p>\n\n\n\n

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

A.P will be x, 100 + x, 200 + x\u2026\u2026\u2026\u2026\u2026\u2026..<\/p>\n\n\n\n

Where, x is first term and<\/p>\n\n\n\n

Common difference, d = 100 + x \u2013 x = 100<\/p>\n\n\n\n

We know, Sn<\/sub> is the sum of n terms of an A.P<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

where, a is first term, d is common difference and n is number of terms in an A.P.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sn<\/sub> = 16500 and n = 10<\/p>\n\n\n\n

S10<\/sub> = 10\/2 [2x + (10 \u2013 1)100]<\/p>\n\n\n\n

16500 = 5{2x + 9(100)}<\/p>\n\n\n\n

16500 = 5(2x + 900)<\/p>\n\n\n\n

16500 = 10x + 4500<\/p>\n\n\n\n

-10x = 4500 \u2013 16500<\/p>\n\n\n\n

\u201310x = \u201312000<\/p>\n\n\n\n

x = 12000\/10<\/p>\n\n\n\n

= 1200<\/p>\n\n\n\n

Hence, his saving in the first year is \u20b9 1200.<\/p>\n\n\n\n

2. A man saves \u20b9 32 during the first year, \u20b9 36 in the second year and in this way he increases his savings by \u20b9 4 every year. Find in what time his saving will be \u20b9 200.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

First year savings is \u20b9 32<\/p>\n\n\n\n

Second year savings is \u20b9 36<\/p>\n\n\n\n

In this process he increases his savings by \u20b9 4 every year<\/p>\n\n\n\n

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

A.P. will be 32, 36, 40,\u2026\u2026\u2026<\/p>\n\n\n\n

Where, 32 is first term and common difference, d = 36 \u2013 32 = 4<\/p>\n\n\n\n

We know, Sn<\/sub> is the sum of n terms of an A.P<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

where, a is first term, d is common difference and n is number of terms in an A.P.<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sn<\/sub> = 200, a = 32, d = 4<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

200 = n\/2 [2(32) + (n-1)4]<\/p>\n\n\n\n

200 = n\/2 [64 + 4n \u2013 4]<\/p>\n\n\n\n

400 = n [60 + 4n]<\/p>\n\n\n\n

400 = 4n [15 + n]<\/p>\n\n\n\n

400\/4 = n [15 + n]<\/p>\n\n\n\n

100 = 15n + n2<\/sup><\/p>\n\n\n\n

n2<\/sup> + 15n \u2013 100 = 0<\/p>\n\n\n\n

n2<\/sup> + 20n \u2013 5n \u2013 100 = 0<\/p>\n\n\n\n

n (n + 20) \u2013 5 (n + 20) = 0<\/p>\n\n\n\n

(n + 20) \u2013 5 (n + 20) = 0<\/p>\n\n\n\n

(n + 20) (n \u2013 5) = 0<\/p>\n\n\n\n

n = -20 or 5<\/p>\n\n\n\n

n = 5 [Since, n is a positive integer]<\/p>\n\n\n\n

Hence, the man requires 5 days to save \u20b9 200<\/p>\n\n\n\n

3. A man arranges to pay off a debt of \u20b9 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the instalment.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

40 annual instalments which form an arithmetic series.<\/p>\n\n\n\n

Let the first instalment be \u2018a\u2019<\/p>\n\n\n\n

S40<\/sub> = 3600, n = 40<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

3600 = 40\/2 [2a + (40-1)d]<\/p>\n\n\n\n

3600 = 20 [2a + 39d]<\/p>\n\n\n\n

3600\/20 = 2a + 39d<\/p>\n\n\n\n

2a + 39d \u2013 180 = 0 \u2026.. (i)<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Sum of first 30 terms is paid and one third of debt is unpaid.<\/p>\n\n\n\n

So, paid amount = 2\/3 \u00d7 3600 = \u20b9 2400<\/p>\n\n\n\n

Sn<\/sub> = 2400, n = 30<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

2400 = 30\/2 [2a + (30-1)d]<\/p>\n\n\n\n

2400 = 15 [2a + 29d]<\/p>\n\n\n\n

2400\/15 = 2a + 29d<\/p>\n\n\n\n

2a + 29d -160 = 0 \u2026. (ii)<\/p>\n\n\n\n

Now, let us solve equation (i) and (ii) by substitution method, we get<\/p>\n\n\n\n

2a + 39d = 180<\/p>\n\n\n\n

2a = 180 \u2013 39d \u2026 (iii)<\/p>\n\n\n\n

Substitute the value of 2a in equation (ii)<\/p>\n\n\n\n

2a + 29d \u2013 160 = 0<\/p>\n\n\n\n

180 \u2013 39d + 29d \u2013 160 = 0<\/p>\n\n\n\n

20 \u2013 10d = 0<\/p>\n\n\n\n

10d = 20<\/p>\n\n\n\n

d = 20\/10<\/p>\n\n\n\n

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

Substitute the value of d in equation (iii)<\/p>\n\n\n\n

2a = 180 \u2013 39d<\/p>\n\n\n\n

2a = 180 \u2013 39(2)<\/p>\n\n\n\n

2a = 180 \u2013 78<\/p>\n\n\n\n

2a = 102<\/p>\n\n\n\n

a = 102\/2<\/p>\n\n\n\n

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

Hence, value of first installment \u2018a\u2019 is \u20b9 51<\/p>\n\n\n\n

4. A manufacturer of the radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find
(\u0456) the production in the first year
(\u0456\u0456) the total product in the 7 years and
(\u0456\u0456\u0456) the product in the 10th<\/sup> year.<\/strong><\/p>\n\n\n\n

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

Given:<\/p>\n\n\n\n

600 and 700 radio sets units are produced in third and seventh year respectively.<\/p>\n\n\n\n

a3<\/sub> = 600 and a7<\/sub> = 700<\/p>\n\n\n\n

(\u0456) <\/strong>The production in the first year<\/p>\n\n\n\n

We need to find the production in the first year.<\/p>\n\n\n\n

Let first year production be \u2018a\u2019<\/p>\n\n\n\n

So the AP formed is, a, a+x, a+2x, \u2026.<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

a3<\/sub> = a + (3-1)d<\/p>\n\n\n\n

600 = a + 2d \u2026. (i)<\/p>\n\n\n\n

a7<\/sub> = a + (7-1)d<\/p>\n\n\n\n

700 = a + 6d<\/p>\n\n\n\n

a = 700 \u2013 6d\u2026. (ii)<\/p>\n\n\n\n

Substitute value of a in (i) we get,<\/p>\n\n\n\n

600 = a + 2d<\/p>\n\n\n\n

600 = 700 \u2013 6d + 2d<\/p>\n\n\n\n

700 \u2013 600 = 4d<\/p>\n\n\n\n

100 = 4d<\/p>\n\n\n\n

d = 100\/4<\/p>\n\n\n\n

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

Now substitute value of d in (ii) we get,<\/p>\n\n\n\n

a = 700 \u2013 6d<\/p>\n\n\n\n

= 700 \u2013 6(25)<\/p>\n\n\n\n

= 700 \u2013 150<\/p>\n\n\n\n

= 550<\/p>\n\n\n\n

\u2234 The production in the first year, \u2018a\u2019 is 550<\/p>\n\n\n\n

(\u0456\u0456) <\/strong>the total product in the 7 years<\/p>\n\n\n\n

We need to find the total product in 7 years i.e. is S7<\/sub><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n-1)d]<\/p>\n\n\n\n

n = 7, a = 550, d = 25<\/p>\n\n\n\n

S7<\/sub> = 7\/2 [2(550) + (7-1)25]<\/p>\n\n\n\n

= 7\/2 [1100 + 150]<\/p>\n\n\n\n

= 7\/2 [1250]<\/p>\n\n\n\n

= 7 [625]<\/p>\n\n\n\n

= 4375<\/p>\n\n\n\n

\u2234 The total product in the 7 years is 4375.<\/p>\n\n\n\n

(\u0456\u0456\u0456) <\/strong>the product in the 10th<\/sup> year.<\/p>\n\n\n\n

We need to find the product in the 10th<\/sup> year i.e. a10<\/sub><\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

an<\/sub> = a + (n-1)d<\/p>\n\n\n\n

n = 10, a = 550, d = 25<\/p>\n\n\n\n

a10<\/sub> = 550 + (10-1)25<\/p>\n\n\n\n

= 550 + (9)25<\/p>\n\n\n\n

= 550 + 225<\/p>\n\n\n\n

= 775<\/p>\n\n\n\n

\u2234 The product in the 10th<\/sup> year is 775.<\/p>\n\n\n\n

5. There are 25 trees at equal distances of 5 meters in a line with a well, the distance of well from the nearest tree being 10 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.<\/strong><\/p>\n\n\n\n

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

Given: total trees are 25 and equal distance between two adjacent trees are 5 meters<\/p>\n\n\n\n

We need to find the total distance the gardener will cover.<\/p>\n\n\n\n

As gardener is coming back to well after watering every tree:<\/p>\n\n\n\n

Distance covered by gardener to water 1st<\/sup> tree and return back to the initial position is 10m + 10m = 20m<\/p>\n\n\n\n

Now, distance between adjacent trees is 5m.<\/p>\n\n\n\n

Distance covered by him to water 2nd<\/sup> tree and return back to the initial position is 15m + 15m = 30m<\/p>\n\n\n\n

Distance covered by the gardener to water 3rd<\/sup> tree return back to the initial position is 20m + 20m = 40m<\/p>\n\n\n\n

Hence distance covered by the gardener to water the trees are in A.P<\/p>\n\n\n\n

A.P. is 20, 30, 40 \u2026\u2026\u2026upto 25 terms<\/p>\n\n\n\n

Here, first term, a = 20, common difference, d = 30 \u2013 20 = 10, n = 25<\/p>\n\n\n\n

We need to find S25<\/sub> which will be the total distance covered by gardener to water 25 trees.<\/p>\n\n\n\n

So by using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

S25<\/sub> = 25\/2 [2(20) + (25-1)10]<\/p>\n\n\n\n

= 25\/2 [40 + (24)10]<\/p>\n\n\n\n

= 25\/2 [40 + 240]<\/p>\n\n\n\n

= 25\/2 [280]<\/p>\n\n\n\n

= 25 [140]<\/p>\n\n\n\n

= 3500<\/p>\n\n\n\n

\u2234 The total distance covered by gardener to water trees all 25 trees is 3500m.<\/p>\n\n\n\n

6. A man is employed to count \u20b9 10710. He counts at the rate of \u20b9 180 per minute for half an hour. After this he counts at the rate of \u20b9 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.<\/strong><\/p>\n\n\n\n

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

Given: Amount to be counted is \u20b9 10710<\/p>\n\n\n\n

We need to find time taken by man to count the entire amount.<\/p>\n\n\n\n

He counts at the rate of \u20b9 180 per minute for half an hour or 30 minutes.<\/p>\n\n\n\n

So, Amount to be counted in an hour = 180 \u00d7 30 = \u20b9 5400<\/p>\n\n\n\n

Amount left = 10710 \u2013 5400 = \u20b9 5310<\/p>\n\n\n\n

Sn<\/sub> = 5310<\/p>\n\n\n\n

After an hour, rate of counting is decreasing at \u20b9 3 per minute. This rate will form an A.P.<\/p>\n\n\n\n

A.P. is 177, 174, 171,\u2026\u2026<\/p>\n\n\n\n

Here a = 177 and d = 174 \u2013 177 = \u20133<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

5310 = n\/2 [2(177) + (n-1) (-3)]<\/p>\n\n\n\n

5310 = n\/2 [354 -3n + 3]<\/p>\n\n\n\n

5310 \u00d7 2 = n [357 \u2013 3n]<\/p>\n\n\n\n

10620 = 357n \u2013 3n2<\/sup><\/p>\n\n\n\n

10620 = 3n(119 \u2013 n)<\/p>\n\n\n\n

10620\/3 = n(119 \u2013 n)<\/p>\n\n\n\n

3540 = 119n \u2013 n2<\/sup><\/p>\n\n\n\n

n2<\/sup> \u2013 119n + 3540 = 0<\/p>\n\n\n\n

n2<\/sup> \u2013 59n \u2013 60n + 3540 = 0<\/p>\n\n\n\n

n(n \u2013 59) \u2013 60(n \u2013 59) = 0<\/p>\n\n\n\n

(n \u2013 59) (n \u2013 60) = 0<\/p>\n\n\n\n

n = 59 or 60<\/p>\n\n\n\n

We shall consider value of n = 59. Since, at 60th<\/sup> min he will count \u20b9 0<\/p>\n\n\n\n

\u2234 The total time taken by him to count the entire amount = 30 + 59 = 89 minutes.<\/p>\n\n\n\n

7. A piece of equipment cost a certain factory \u20b9 600,000. If it depreciates in value 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?<\/strong><\/p>\n\n\n\n

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

Given: A piece of equipment cost a certain factory is \u20b9 600,000<\/p>\n\n\n\n

We need to find the value of the equipment at the end of 10 years.<\/p>\n\n\n\n

The price of equipment depreciates 15%, 13.5%, 12% in 1st<\/sup>, 2nd<\/sup>, 3rd<\/sup> year and so on.<\/p>\n\n\n\n

So the A.P. will be 15, 13.5, 12,\u2026\u2026\u2026\u2026\u2026 up to 10 terms<\/p>\n\n\n\n

Here, a = 15, d = 13.5 \u2013 15 = \u20131.5, n = 10<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

S10<\/sub> = 10\/2 [2(15) + (10-1) (-1.5)]<\/p>\n\n\n\n

= 5 [30 + 9(-1.5)]<\/p>\n\n\n\n

= 5 [30 \u2013 13.5]<\/p>\n\n\n\n

= 5 [16.5]<\/p>\n\n\n\n

= 82.5<\/p>\n\n\n\n

The value of equipment at the end of 10 years is = [100 \u2013 Depreciation %]\/100 \u00d7 cost<\/p>\n\n\n\n

= [100 \u2013 82.5]\/100 \u00d7 600000<\/p>\n\n\n\n

= 175\/10 \u00d7 6000<\/p>\n\n\n\n

= 175 \u00d7 600<\/p>\n\n\n\n

= 105000<\/p>\n\n\n\n

\u2234 The value of equipment at the end of 10 years is \u20b9 105000.<\/p>\n\n\n\n

8. A farmer buys a used tractor for \u20b9 12000. He pays \u20b9 6000 cash and agrees to pay the balance in annual instalments of \u20b9 500 plus 12% interest on the unpaid amount. How much the tractor cost him?<\/strong><\/p>\n\n\n\n

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

Given: Price of the tractor is \u20b912000.<\/p>\n\n\n\n

We need to find the total cost of the tractor if he buys it in installments.<\/p>\n\n\n\n

Total price = \u20b9 12000<\/p>\n\n\n\n

Paid amount = \u20b9 6000<\/p>\n\n\n\n

Unpaid amount = \u20b9 12000 \u2013 6000 = \u20b9 6000<\/p>\n\n\n\n

He pays remaining \u20b9 6000 in \u2018n\u2019 number of installments of \u20b9 500 each.<\/p>\n\n\n\n

So, n = 6000\/500 = 12<\/p>\n\n\n\n

Cost incurred by him to pay remaining 6000 is<\/p>\n\n\n\n

The AP will be:<\/p>\n\n\n\n

(500 + 12% of 6000) + (500 + 12% of 5500) + \u2026 up to 12 terms<\/p>\n\n\n\n

500 \u00d7 12 + 12% of (6000 + 5500 + \u2026 up to 12 terms)<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Sn<\/sub> = n\/2 [2a + (n \u2013 1)d]<\/p>\n\n\n\n

n = 12, a = 6000, d = -500<\/p>\n\n\n\n

S12<\/sub> = 500\u00d712 + 12\/100 \u00d7 12\/2 [2(6000) + (12-1) (-500)]<\/p>\n\n\n\n

= 6000 + 72\/100 [12000 + 11 (-500)]<\/p>\n\n\n\n

= 6000 + 72\/100 [12000 \u2013 5500]<\/p>\n\n\n\n

= 6000 + 72\/100 [6500]<\/p>\n\n\n\n

= 6000 + 4680<\/p>\n\n\n\n

= 10680<\/p>\n\n\n\n

Total cost = 6000 + 10680<\/p>\n\n\n\n

= 16680<\/p>\n\n\n\n

\u2234 The total cost of the tractor if he buys it in installment is \u20b9 16680.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 19: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 19\u2013Arithmetic Progressions<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 19\u2013Arithmetic Progressions PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 11 Maths :<\/strong><\/h2>\n\n\n\n