{"id":55173,"date":"2020-11-20T12:19:45","date_gmt":"2020-11-20T12:19:45","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=55173"},"modified":"2023-09-19T02:24:09","modified_gmt":"2023-09-19T02:24:09","slug":"ncert-solutions-for-class-10th-mathematics-chapter-5-arithmetic-progressions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-5-arithmetic-progressions\/","title":{"rendered":"NCERT Solutions for Class 10th Mathematics: Chapter 5 – Arithmetic Progressions"},"content":{"rendered":"\n

Class 10: Mathematics Chapter 5 solutions. Complete Class 10 Mathematics Chapter 5 Notes.<\/p>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 5 – Arithmetic Progressions<\/strong><\/h2>\n\n\n\n

NCERT 10th Mathematics Chapter 5, class 10 Mathematics Chapter 5 solutions<\/p>\n\n\n\n

Page No: 99<\/p>\n\n\n\n

Exercise 5.1<\/h4>\n\n\n\n

1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?<\/strong><\/p>\n\n\n\n

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39<\/p>\n\n\n\n

Clearly 15, 23, 31, 39 \u2026 forms an A.P. because every term is 8 more than the preceding term.<\/p>\n\n\n\n

(ii) The amount of air present in a cylinder when a vacuum pump removes 1\/4 of the air remaining in the cylinder at a time.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Let the initial volume of air in a cylinder be V<\/em> litres. In each stroke, the vacuum pump removes 1\/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 – 1\/4 = 3\/4th part of air will remain.
Therefore, volumes will be V<\/em>, 3V<\/em>\/4 , (3V<\/em>\/4)2<\/sup> , (3V<\/em>\/4)3<\/sup>…
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.<\/p>\n\n\n\n

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 \u2026 forms an A.P. because every term is 50 more than the preceding term.<\/p>\n\n\n\n

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
We know that if Rs P<\/em> is deposited at r<\/em>% compound interest per annum for n years, our money will be<\/p>\n\n\n\n

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

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

2. Write first four terms of the A.P. when the first term a and the common differenced are given as follows
<\/strong>(i) a<\/em> = 10, d<\/em> = 10
(ii) a<\/em> = -2, d<\/em> = 0
(iii) a<\/em> = 4, d<\/em> = – 3
(iv) a<\/em> = -1 d<\/em> = 1\/2
(v) a<\/em> = – 1.25, d<\/em> = – 0.25<\/p>\n\n\n\n

Answer<\/strong>
(i) a<\/em> = 10, d<\/em> = 10
Let the series be a<\/em>1<\/sub>, a<\/em>2<\/sub>, a<\/em>3<\/sub>, a<\/em>4<\/sub>, a<\/em>5<\/sub> \u2026
a<\/em>1<\/sub> = a<\/em> = 10
a<\/em>2<\/sub> = a<\/em>1<\/sub> + d<\/em> = 10 + 10 = 20
a<\/em>3<\/sub> = a<\/em>2<\/sub> + d<\/em> = 20 + 10 = 30
a<\/em>4<\/sub> = a<\/em>3<\/sub> + d<\/em> = 30 + 10 = 40
a<\/em>5<\/sub> = a<\/em>4<\/sub> + d<\/em> = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 \u2026
First four terms of this A.P. will be 10, 20, 30, and 40.<\/p>\n\n\n\n

(ii) a<\/em> = – 2, d<\/em> = 0
Let the series be a<\/em>1<\/sub>, a2<\/sub>, a<\/em>3<\/sub>, a<\/em>4<\/sub> \u2026
a<\/em>1<\/sub> = a<\/em> = -2
a<\/em>2<\/sub> = a<\/em>1<\/sub> + d<\/em> = – 2 + 0 = – 2
a<\/em>3<\/sub> = a<\/em>2<\/sub> + d = – 2 + 0 = – 2
a<\/em>4<\/sub> = a<\/em>3<\/sub> + d<\/em> = – 2 + 0 = – 2
Therefore, the series will be – 2, – 2, – 2, – 2 \u2026
First four terms of this A.P. will be – 2, – 2, – 2 and – 2.<\/p>\n\n\n\n

(iii) a<\/em> = 4, d<\/em> = – 3
Let the series be a<\/em>1<\/sub>, a<\/em>2<\/sub>, a<\/em>3<\/sub>, a<\/em>4<\/sub> \u2026
a<\/em>1<\/sub> = a<\/em> = 4
a<\/em>2<\/sub> = a<\/em>1<\/sub> + d<\/em> = 4 – 3 = 1
a<\/em>3<\/sub> = a<\/em>2<\/sub> + d<\/em> = 1 – 3 = – 2
a<\/em>4<\/sub> = a<\/em>3<\/sub> + d<\/em> = – 2 – 3 = – 5
Therefore, the series will be 4, 1, – 2 – 5 \u2026
First four terms of this A.P. will be 4, 1, – 2 and – 5.<\/p>\n\n\n\n

(iv) a<\/em> = – 1, d<\/em> = 1\/2
Let the series be a<\/em>1<\/sub>, a<\/em>2<\/sub>, a<\/em>3<\/sub>, a<\/em>4<\/sub> \u2026a<\/em>1<\/sub> = a<\/em> = -1
a<\/em>2<\/sub> = a<\/em>1<\/sub> + d<\/em> = -1 + 1\/2 = -1\/2
a<\/em>3<\/sub> = a<\/em>2<\/sub> + d<\/em> = -1\/2 + 1\/2 = 0
a<\/em>4<\/sub> = a<\/em>3<\/sub> + d<\/em> = 0 + 1\/2 = 1\/2
Clearly, the series will be-1, -1\/2, 0, 1\/2
First four terms of this A.P. will be -1, -1\/2, 0 and 1\/2.<\/p>\n\n\n\n

(v) a<\/em> = – 1.25, d<\/em> = – 0.25
Let the series be a<\/em>1<\/sub>, a<\/em>2<\/sub>, a<\/em>3<\/sub>, a<\/em>4<\/sub> \u2026
a<\/em>1<\/sub> = a<\/em> = – 1.25
a<\/em>2<\/sub> = a<\/em>1<\/sub> + d<\/em> = – 1.25 – 0.25 = – 1.50
a<\/em>3<\/sub> = a<\/em>2<\/sub> + d<\/em> = – 1.50 – 0.25 = – 1.75
a<\/em>4<\/sub> = a<\/em>3<\/sub> + d<\/em> = – 1.75 – 0.25 = – 2.00
Clearly, the series will be 1.25, – 1.50, – 1.75, – 2.00 \u2026\u2026..
First four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

3. For the following A.P.s, write the first term and the common difference.
<\/strong>(i) 3, 1, – 1, – 3 \u2026
(ii) -5, – 1, 3, 7 \u2026
(iii) 1\/3, 5\/3, 9\/3, 13\/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 \u2026<\/p>\n\n\n\n

Answer<\/strong>
(i) 3, 1, – 1, – 3 \u2026
Here, first term, a<\/em> = 3
Common difference, d<\/em> = Second term – First term
= 1 – 3 = – 2<\/p>\n\n\n\n

(ii) – 5, – 1, 3, 7 \u2026
Here, first term, a<\/em> = – 5
Common difference, d<\/em> = Second term – First term
= ( – 1) – ( – 5) = – 1 + 5 = 4
(iii) 1\/3, 5\/3, 9\/3, 13\/3 ….
Here, first term, a<\/em> = 1\/3<\/p>\n\n\n\n

Common difference, d<\/em> = Second term – First term <\/p>\n\n\n\n

= 5\/3 – 1\/3 = 4\/3<\/p>\n\n\n\n

(iv) 0.6, 1.7, 2.8, 3.9 \u2026
Here, first term, a<\/em> = 0.6
Common difference, d<\/em> = Second term – First term
= 1.7 – 0.6
= 1.1<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

4. Which of the following are APs? If they form an A.P. find the common difference d<\/em> and write three more terms.
<\/strong>(i) 2, 4, 8, 16 \u2026
(ii) 2, 5\/2, 3, 7\/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 \u2026
(iv) -10, – 6, – 2, 2 \u2026
(v) 3, 3 + \u221a2, 3 + 2\u221a2, 3 + 3\u221a2
(vi) 0.2, 0.22, 0.222, 0.2222 \u2026.
(vii) 0, – 4, – 8, – 12 \u2026
(viii) -1\/2, -1\/2, -1\/2, -1\/2 ….
(ix) 1, 3, 9, 27 \u2026
(x) a<\/em>, 2a<\/em>, 3a<\/em>, 4a<\/em> \u2026
(xi) a<\/em>, a<\/em>2<\/sup>, a<\/em>3<\/sup>, a<\/em>4<\/sup> \u2026
(xii) \u221a2, \u221a8, \u221a18, \u221a32 …
(xiii) \u221a3, \u221a6, \u221a9, \u221a12 …
(xiv) 12<\/sup>, 32<\/sup>, 52<\/sup>, 72<\/sup> \u2026
(xv) 12<\/sup>, 52<\/sup>, 72<\/sup>, 73<\/sup> \u2026<\/p>\n\n\n\n

Answer<\/strong>
(i) 2, 4, 8, 16 \u2026
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = 4 – 2 = 2
a<\/em>3<\/sub> – a<\/em>2<\/sub> = 8 – 4 = 4
a<\/em>4<\/sub> – a<\/em>3<\/sub> = 16 – 8 = 8
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is not the same every time.<\/p>\n\n\n\n

Therefore, the given numbers are forming an A.P.<\/p>\n\n\n\n

(ii) 2, 5\/2, 3, 7\/2 ….
Here,<\/p>\n\n\n\n

a<\/em>2<\/sub> – a<\/em>1<\/sub> = 5\/2 – 2 = 1\/2
a<\/em>3<\/sub> – a<\/em>2<\/sub> = 3 – 5\/2 = 1\/2
a<\/em>4<\/sub> – a<\/em>3<\/sub> = 7\/2 – 3 = 1\/2
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is same every time.
Therefore, d<\/em> = 1\/2 and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = 7\/2 + 1\/2 = 4
a<\/em>6<\/sub> = 4 + 1\/2 = 9\/2
a<\/em>7<\/sub> = 9\/2 + 1\/2 = 5<\/p>\n\n\n\n

(iii) -1.2, – 3.2, -5.2, -7.2 \u2026
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = ( -3.2) – ( -1.2) = -2
a<\/em>3<\/sub> – a<\/em>2<\/sub> = ( -5.2) – ( -3.2) = -2
a<\/em>4<\/sub> – a<\/em>3<\/sub> = ( -7.2) – ( -5.2) = -2
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is same every time.
Therefore, d<\/em> = -2 and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = – 7.2 – 2 = – 9.2
a<\/em>6<\/sub> = – 9.2 – 2 = – 11.2
a<\/em>7<\/sub> = – 11.2 – 2 = – 13.2<\/p>\n\n\n\n

(iv) -10, – 6, – 2, 2 \u2026
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = (-6) – (-10) = 4
a<\/em>3<\/sub> – a<\/em>2<\/sub> = (-2) – (-6) = 4
a<\/em>4<\/sub> – a<\/em>3<\/sub> = (2) – (-2) = 4
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is same every time.
Therefore, d<\/em> = 4 and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = 2 + 4 = 6
a<\/em>6<\/sub> = 6 + 4 = 10
a<\/em>7<\/sub> = 10 + 4 = 14<\/p>\n\n\n\n

(v) 3, 3 + \u221a2, 3 + 2\u221a2, 3 + 3\u221a2
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = 3 + \u221a2 – 3 = \u221a2
a<\/em>3<\/sub> – a<\/em>2<\/sub> = (3 + 2\u221a2) – (3 + \u221a2) = \u221a2
a<\/em>4<\/sub> – a<\/em>3<\/sub> = (3 + 3\u221a2) – (3 + 2\u221a2) = \u221a2
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is same every time.
Therefore, d<\/em> = \u221a2 and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = (3 + \u221a2) + \u221a2 = 3 + 4\u221a2
a<\/em>6<\/sub> = (3 + 4\u221a2) + \u221a2 = 3 + 5\u221a2
a<\/em>7<\/sub> = (3 + 5\u221a2) + \u221a2 = 3 + 6\u221a2<\/p>\n\n\n\n

(vi) 0.2, 0.22, 0.222, 0.2222 \u2026.
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = 0.22 – 0.2 = 0.02
a<\/em>3<\/sub> – a<\/em>2<\/sub> = 0.222 – 0.22 = 0.002
a<\/em>4<\/sub> – a<\/em>3<\/sub> = 0.2222 – 0.222 = 0.0002
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is not the same every time.<\/p>\n\n\n\n

Therefore, the given numbers are forming an A.P.<\/p>\n\n\n\n

(vii) 0, -4, -8, -12 \u2026
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = (-4) – 0 = -4
a<\/em>3<\/sub> – a<\/em>2<\/sub> = (-8) – (-4) = -4
a<\/em>4<\/sub> – a<\/em>3<\/sub> = (-12) – (-8) = -4
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is same every time.
Therefore, d<\/em> = -4 and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = -12 – 4 = -16
a<\/em>6<\/sub> = -16 – 4 = -20
a<\/em>7<\/sub> = -20 – 4 = -24<\/p>\n\n\n\n

(viii) -1\/2, -1\/2, -1\/2, -1\/2 ….
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = (-1\/2) – (-1\/2) = 0
a<\/em>3<\/sub> – a<\/em>2<\/sub> = (-1\/2) – (-1\/2) = 0
a<\/em>4<\/sub> – a<\/em>3<\/sub> = (-1\/2) – (-1\/2) = 0
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is same every time.
Therefore, d<\/em> = 0 and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = (-1\/2) – 0 = -1\/2
a<\/em>6<\/sub> = (-1\/2) – 0 = -1\/2
a<\/em>7<\/sub> = (-1\/2) – 0 = -1\/2<\/p>\n\n\n\n

(ix) 1, 3, 9, 27 \u2026
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = 3 – 1 = 2
a<\/em>3<\/sub> – a<\/em>2<\/sub> = 9 – 3 = 6
a<\/em>4<\/sub> – a<\/em>3<\/sub> = 27 – 9 = 18
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is not the same every time.<\/p>\n\n\n\n

Therefore, the given numbers are forming an A.P.<\/p>\n\n\n\n

(x) a<\/em>, 2a<\/em>, 3a<\/em>, 4a<\/em> \u2026
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = 2a<\/em> – a <\/em>= a<\/em>
a<\/em>3<\/sub> – a<\/em>2<\/sub> = 3a<\/em> – 2a<\/em> = a<\/em>
a<\/em>4<\/sub> – a<\/em>3<\/sub> = 4a<\/em> – 3a<\/em> = a<\/em>
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is same every time.
Therefore, d<\/em> = a<\/em> and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = 4a<\/em> + a<\/em> = 5a<\/em>
a<\/em>6<\/sub> = 5a <\/em>+ a<\/em> = 6a<\/em>
a<\/em>7<\/sub> = 6a<\/em> + a<\/em> = 7a<\/em><\/p>\n\n\n\n

(xi) a<\/em>, a<\/em>2<\/sup>, a<\/em>3<\/sup>, a<\/em>4<\/sup> \u2026
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = a<\/em>2 <\/sup>– a<\/em> = (a<\/em> – 1)
a<\/em>3<\/sub> – a<\/em>2<\/sub> = a<\/em>3 <\/sup>–a<\/em>2 <\/sup>= a<\/em>2 <\/sup>(a<\/em> – 1)
a<\/em>4<\/sub> – a<\/em>3<\/sub> = a<\/em>4<\/sup> – a<\/em>3 <\/sup>= a<\/em>3<\/sup>(a<\/em> – 1)
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is not the same every time.<\/p>\n\n\n\n

Therefore, the given numbers are forming an A.P.<\/p>\n\n\n\n

(xii) \u221a2, \u221a8, \u221a18, \u221a32 …
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = \u221a8 – \u221a2  = 2\u221a2 – \u221a2 = \u221a2
a<\/em>3<\/sub> – a<\/em>2<\/sub> = \u221a18 – \u221a8 = 3\u221a2 – 2\u221a2 = \u221a2
a<\/em>4<\/sub> – a<\/em>3<\/sub> = 4\u221a2 – 3\u221a2 = \u221a2
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is same every time.
Therefore, d<\/em> = \u221a2 and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = \u221a32  + \u221a2 = 4\u221a2 + \u221a2 = 5\u221a2 = \u221a50
a<\/em>6<\/sub> = 5\u221a2 +\u221a2 = 6\u221a2 = \u221a72
a<\/em>7<\/sub> = 6\u221a2 + \u221a2 = 7\u221a2 = \u221a98<\/p>\n\n\n\n


(xiii) \u221a3, \u221a6, \u221a9, \u221a12 …
Here,
a<\/em>2<\/sub> – a<\/em>1<\/sub> = \u221a6 – \u221a3 = \u221a3 \u00d7 2 -\u221a3 = \u221a3(\u221a2 – 1)
a<\/em>3<\/sub> – a<\/em>2<\/sub> = \u221a9 – \u221a6 = 3 – \u221a6 = \u221a3(\u221a3 – \u221a2)
a<\/em>4<\/sub> – a<\/em>3<\/sub> = \u221a12 – \u221a9 = 2\u221a3 – \u221a3 \u00d7 3 = \u221a3(2 – \u221a3)
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is not the same every time.<\/p>\n\n\n\n

Therefore, the given numbers are forming an A.P.<\/p>\n\n\n\n

(xiv) 12<\/sup>, 32<\/sup>, 52<\/sup>, 72<\/sup> \u2026<\/p>\n\n\n\n

Or, 1, 9, 25, 49 \u2026..
Here,
a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 9 \u2212 1 = 8
a<\/em>3<\/sub> \u2212 a<\/em>2 <\/sub>= 25 \u2212 9 = 16
a<\/em>4<\/sub> \u2212 a<\/em>3<\/sub> = 49 \u2212 25 = 24
\u21d2 a<\/em>n<\/em>+1<\/sub> – a<\/em>n<\/em><\/sub> is not the same every time.<\/p>\n\n\n\n

Therefore, the given numbers are forming an A.P.<\/p>\n\n\n\n

(xv) 12<\/sup>, 52<\/sup>, 72<\/sup>, 73 \u2026
Or 1, 25, 49, 73 \u2026
Here,
a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 25 \u2212 1 = 24
a<\/em>3<\/sub> \u2212 a<\/em>2 <\/sub>= 49 \u2212 25 = 24
a<\/em>4<\/sub> \u2212 a<\/em>3<\/sub> = 73 \u2212 49 = 24
i.e., a<\/em>k<\/em>+1 <\/sub>\u2212 a<\/em>k<\/em><\/sub> is same every time.
\u21d2 a<\/em>n+1<\/sub> – a<\/em>n<\/sub> is same every time.
Therefore, d<\/em> = 24 and the given numbers are in A.P.
Three more terms are
a<\/em>5<\/sub> = 73+ 24 = 97
a<\/em>6<\/sub> = 97 + 24 = 121
a<\/em>7 <\/sub>= 121 + 24 = 145<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

Page No: 105<\/p>\n\n\n\n

Exercise 5.2<\/h5>\n\n\n\n

1. Fill in the blanks in the following table, given that a<\/em> is the first term, d<\/em> the common difference and a<\/em>n<\/em><\/sub> the n<\/em>th<\/sup> term of the A.P.<\/strong>
<\/p>\n\n\n\n

<\/td>a<\/em><\/td>d<\/em><\/td>n<\/em><\/td>a<\/em>n<\/em><\/sub><\/td><\/tr>
(i)<\/td>7<\/td>3<\/td>8<\/td>\u2026…<\/td><\/tr>
(ii)<\/td>\u2212 18<\/td>\u2026..<\/td>10<\/td>0<\/td><\/tr>
(iii)<\/td>\u2026..<\/td>\u2212 3<\/td>18<\/td>\u2212 5<\/td><\/tr>
(iv)<\/td>\u2212 18.9<\/td>2.5<\/td>\u2026..<\/td>3.6<\/td><\/tr>
(v)<\/td>3.5<\/td>0<\/td>105<\/td>\u2026..<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Answer<\/strong>
(i) a<\/em> = 7, d<\/em> = 3, n<\/em> = 8, a<\/em>n<\/em><\/sub> = ?
We know that,
For an A.P. a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
= 7 + (8 \u2212 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence, a<\/em>n<\/em><\/sub> = 28<\/p>\n\n\n\n

(ii) Given that
a<\/em> = \u221218, n<\/em> = 10, a<\/em>n<\/em><\/sub> = 0, d<\/em> = ?
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
0 = \u2212 18 + (10 \u2212 1) d<\/em>
18 = 9d<\/em>
d<\/em> = 18\/9 = 2
Hence, common difference, d <\/em>= 2<\/p>\n\n\n\n

(iii) Given that
d <\/em>= \u22123, n<\/em> = 18, a<\/em>n<\/em><\/sub> = \u22125
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
\u22125 = a<\/em> + (18 \u2212 1) (\u22123)
\u22125 = a<\/em> + (17) (\u22123)
\u22125 = a <\/em>\u2212 51
a<\/em> = 51 \u2212 5 = 46
Hence, a<\/em> = 46<\/p>\n\n\n\n

(iv) a<\/em> = \u221218.9, d<\/em> = 2.5, a<\/em>n<\/em><\/sub> = 3.6, n<\/em> = ?
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
3.6 = \u2212 18.9 + (n<\/em> \u2212 1) 2.5
3.6 + 18.9 = (n<\/em> \u2212 1) 2.5
22.5 = (n<\/em> \u2212 1) 2.5
(n<\/em> – 1) = 22.5\/2.5
n<\/em> – 1 = 9
n<\/em> = 10
Hence, n<\/em> = 10<\/p>\n\n\n\n

(v) a<\/em> = 3.5, d<\/em> = 0, n<\/em> = 105, a<\/em>n<\/em><\/sub> = ?
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>n<\/em><\/sub> = 3.5 + (105 \u2212 1) 0
a<\/em>n<\/em><\/sub> = 3.5 + 104 \u00d7 0
a<\/em>n<\/em><\/sub> = 3.5
Hence, a<\/em>n<\/em><\/sub> = 3.5<\/p>\n\n\n\n

Choose the correct choice in the following and justify
(i) 30th<\/sup> term of the A.P: 10, 7, 4, \u2026, is
(A)97 (B)77 (C)\u221277 (D.)\u221287

(ii) 11th <\/sup>term of the A.P. -3, -1\/2, ,2 …. is
(A) 28 (B) 22 (C) – 38 (D)<\/a><\/p>\n\n\n\n

Answer<\/strong>
(i) Given that
A.P. 10, 7, 4, \u2026
First term, a<\/em> = 10
Common difference, d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1 <\/sub>= 7 \u2212 10 = \u22123
We know that, a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>30<\/sub> = 10 + (30 \u2212 1) (\u22123)
a<\/em>30<\/sub> = 10 + (29) (\u22123)
a<\/em>30<\/sub> = 10 \u2212 87 = \u221277
Hence, the correct answer is option C.<\/p>\n\n\n\n

(ii) Given that A.P. is -3, -1\/2, ,2 …
First term a<\/em> = – 3
Common difference, d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = (-1\/2) – (-3)
= (-1\/2) + 3 = 5\/2
We know that,a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>11<\/sub> = 3 + (11 -1)(5\/2)
a<\/em>11<\/sub> = 3 + (10)(5\/2)
a<\/em>11<\/sub> = -3 + 25
a<\/em>11<\/sub> = 22
Hence, the answer is option B.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

3. In the following APs find the missing term in the boxes.<\/strong><\/p>\n\n\n\n

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

Answer<\/strong>
(i) For this A.P.,<\/p>\n\n\n\n

a<\/em> = 2
a<\/em>3<\/sub> = 26
We know that, a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>3<\/sub> = 2 + (3 – 1) d<\/em>
26 = 2 + 2d<\/em>
24 = 2d<\/em>
d<\/em> = 12
a<\/em>2<\/sub> = 2 + (2 – 1) 12
= 14
Therefore, 14 is the missing term.<\/p>\n\n\n\n

(ii) For this A.P.,
a<\/em>2<\/sub> = 13 and
a<\/em>4<\/sub> = 3
We know that, a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>2<\/sub> = a<\/em> + (2 – 1) d<\/em>
13 = a<\/em> + d<\/em> … (i)<\/strong>
a<\/em>4<\/sub> = a<\/em> + (4 – 1) d<\/em>
3 = a<\/em> + 3d<\/em> … (ii)<\/strong>
On subtracting (i)<\/strong> from (ii)<\/strong>, we get
– 10 = 2d<\/em>
d<\/em> = – 5
From equation (i)<\/strong>, we get
13 = a<\/em> + (-5)
a<\/em> = 18
a<\/em>3<\/sub> = 18 + (3 – 1) (-5)
= 18 + 2 (-5) = 18 – 10 = 8
Therefore, the missing terms are 18 and 8 respectively.<\/p>\n\n\n\n

(iii) For this A.P.,
a <\/em>= 5 and
a<\/em>4<\/sub> = 19\/2
We know that, a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>4<\/sub> = a<\/em> + (4 – 1) d<\/em>
19\/2 = 5<\/em> + 3d
19\/2 – 5 = 3d3d = 9\/2
d = 3\/2<\/p>\n\n\n\n

a<\/em>2<\/sub> = a<\/em> + (2 – 1) d<\/em>
a<\/em>2<\/sub> = 5 <\/em>+ 3\/2
a<\/em>2<\/sub> = 13\/2<\/p>\n\n\n\n

a<\/em>3<\/sub> = a<\/em> + (3 – 1) d<\/em><\/p>\n\n\n\n

a<\/em>3<\/sub> = 5<\/em> + 2\u00d73\/2<\/p>\n\n\n\n

a<\/em>3<\/sub> = 8<\/em><\/p>\n\n\n\n

Therefore, the missing terms are 13\/2 and 8 respectively.<\/p>\n\n\n\n

(iv) For this A.P.,
a<\/em> = \u22124 and
a<\/em>6<\/sub> = 6
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a6<\/sub> = a + (6 \u2212 1) d
6 = \u2212 4 + 5d<\/em>
10 = 5d<\/em>
d<\/em> = 2
a<\/em>2<\/sub> = a<\/em> + d<\/em> = \u2212 4 + 2 = \u22122
a<\/em>3<\/sub> = a<\/em> + 2d<\/em> = \u2212 4 + 2 (2) = 0
a<\/em>4<\/sub> = a<\/em> + 3d<\/em> = \u2212 4 + 3 (2) = 2
a<\/em>5<\/sub> = a <\/em>+ 4d<\/em> = \u2212 4 + 4 (2) = 4
Therefore, the missing terms are \u22122, 0, 2, and 4 respectively.<\/p>\n\n\n\n

(v)
For this A.P.,
a<\/em>2<\/sub> = 38
a<\/em>6<\/sub> = \u221222
We know that
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>2<\/sub> = a<\/em> + (2 \u2212 1) d<\/em>
38 = a<\/em> + d<\/em> … (i)<\/strong>
a<\/em>6<\/sub> = a<\/em> + (6 \u2212 1) d<\/em>
\u221222 = a<\/em> + 5d<\/em> … (ii)<\/strong>
On subtracting equation (i)<\/strong> from (ii)<\/strong>, we get
\u2212 22 \u2212 38 = 4d<\/em>
\u221260 = 4d<\/em>
d<\/em> = \u221215
a<\/em> = a<\/em>2<\/sub> \u2212 d<\/em> = 38 \u2212 (\u221215) = 53
a<\/em>3<\/sub> = a <\/em>+ 2d <\/em>= 53 + 2 (\u221215) = 23
a<\/em>4<\/sub> = a<\/em> + 3d<\/em> = 53 + 3 (\u221215) = 8
a<\/em>5<\/sub> = a<\/em> + 4d<\/em> = 53 + 4 (\u221215) = \u22127
Therefore, the missing terms are 53, 23, 8, and \u22127 respectively.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

4. Which term of the A.P. 3, 8, 13, 18, \u2026 is 78?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
3, 8, 13, 18, \u2026
For this A.P.,
a<\/em> = 3
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 8 \u2212 3 = 5
Let n<\/em>th<\/sup> term of this A.P. be 78.
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
78 = 3 + (n<\/em> \u2212 1) 5
75 = (n<\/em> \u2212 1) 5
(n<\/em> \u2212 1) = 15
n<\/em> = 16
Hence, 16th<\/sup> term of this A.P. is 78.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

5. Find the number of terms in each of the following A.P.<\/strong>
(i) 7, 13, 19, \u2026, 205
(ii) 18,<\/p>\n\n\n\n

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

, 13,…., -47<\/p>\n\n\n\n

Answer<\/strong>
(i) For this A.P.,
a<\/em> = 7
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 13 \u2212 7 = 6
Let there are n<\/em> terms in this A.P.
a<\/em>n<\/em><\/sub> = 205
We know that
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
Therefore, 205 = 7 + (n <\/em>\u2212 1) 6
198 = (n<\/em> \u2212 1) 6
33 = (n<\/em> \u2212 1)
n<\/em> = 34
Therefore, this given series has 34 terms in it.<\/p>\n\n\n\n

(ii) For this A.P.,
a<\/em> = 18

Let there are n terms in this A.P.
a<\/em>n<\/em><\/sub> = 205<\/p>\n\n\n\n

a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em><\/p>\n\n\n\n

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

-47 – 18 = (n<\/em> – 1) (-5\/2)
-65 = (n<\/em> – 1)(-5\/2)
(n<\/em> – 1) = -130\/-5
(n<\/em> – 1) = 26
n <\/em>= 27
Therefore, this given A.P. has 27 terms in it.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, \u2026<\/strong> <\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

For this A.P.,
a<\/em> = 11
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 8 \u2212 11 = \u22123
Let \u2212150 be the n<\/em>th<\/sup> term of this A.P.
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
-150 = 11 + (n<\/em> – 1)(-3)
-150 = 11 – 3n<\/em> + 3
-164 = -3n<\/em>
n<\/em> = 164\/3
Clearly, n<\/em> is not an integer.
Therefore, – 150 is not a term of this A.P.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

7. Find the 31st<\/sup> term of an A.P. whose 11th<\/sup> term is 38 and the 16th<\/sup> term is 73.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given that,
a<\/em>11<\/sub> = 38
a<\/em>16<\/sub> = 73
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>11<\/sub> = a <\/em>+ (11 \u2212 1) d<\/em>
38 = a<\/em> + 10d<\/em> … (i)<\/strong>
Similarly,
a<\/em>16<\/sub> = a<\/em> + (16 \u2212 1) d<\/em>
73 = a<\/em> + 15d<\/em> … (ii)<\/strong>
On subtracting (i)<\/strong> from (ii)<\/strong>, we get
35 = 5d<\/em>
d<\/em> = 7
From equation (i)<\/strong>,
38 = a<\/em> + 10 \u00d7 (7)
38 \u2212 70 = a<\/em>
a<\/em> = \u221232
a<\/em>31<\/sub> = a <\/em>+ (31 \u2212 1) d<\/em>
= \u2212 32 + 30 (7)
= \u2212 32 + 210
= 178
Hence, 31st<\/sup> term is 178.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

8. An A.P. consists of 50 terms of which 3rd<\/sup> term is 12 and the last term is 106. Find the 29th<\/sup> term.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given that,
a<\/em>3<\/sub> = 12
a<\/em>50<\/sub> = 106
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>3<\/sub> = a<\/em> + (3 \u2212 1) d<\/em>
12 = a<\/em> + 2d<\/em> … (i)<\/strong>
Similarly, a<\/em>50 <\/sub>= a<\/em> + (50 \u2212 1) d<\/em>
106 = a<\/em> + 49d<\/em> … (ii)<\/strong>
On subtracting (i)<\/strong> from (ii)<\/strong>, we get
94 = 47d<\/em>
d<\/em> = 2
From equation (i)<\/strong>, we get
12 = a<\/em> + 2 (2)
a<\/em> = 12 \u2212 4 = 8
a<\/em>29<\/sub> = a<\/em> + (29 \u2212 1) d<\/em>
a<\/em>29<\/sub> = 8 + (28)2
a<\/em>29<\/sub> = 8 + 56 = 64
Therefore, 29th<\/sup> term is 64.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

9. If the 3rd<\/sup> and the 9th<\/sup> terms of an A.P. are 4 and \u2212 8 respectively. Which term of this A.P. is zero.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given that,
a<\/em>3<\/sub> = 4
a<\/em>9<\/sub> = \u22128
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>3<\/sub> = a<\/em> + (3 \u2212 1) d<\/em>
4 = a<\/em> + 2d<\/em> … (i)<\/strong>
a<\/em>9<\/sub> = a <\/em>+ (9 \u2212 1) d<\/em>
\u22128 = a<\/em> + 8d<\/em> … (ii)<\/strong>
On subtracting equation (i)<\/strong> from (ii)<\/strong>, we get,
\u221212 = 6d<\/em>
d<\/em> = \u22122
From equation (i)<\/strong>, we get,
4 = a <\/em>+ 2 (\u22122)
4 = a<\/em> \u2212 4
a<\/em> = 8
Let n<\/em>th<\/sup> term of this A.P. be zero.
a<\/em>n <\/em><\/sub>= a <\/em>+ (n <\/em>\u2212 1) d<\/em>
0 = 8 + (n<\/em> \u2212 1) (\u22122)
0 = 8 \u2212 2n<\/em> + 2
2n <\/em>= 10
n<\/em> = 5
Hence, 5th<\/sup> term of this A.P. is 0.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

10. If 17th<\/sup> term of an A.P. exceeds its 10th<\/sup> term by 7. Find the common difference.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
We know that,
For an A.P., a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>17<\/sub> = a<\/em> + (17 \u2212 1) d<\/em>
a<\/em>17<\/sub> = a<\/em> + 16d<\/em>
Similarly, a<\/em>10<\/sub> = a<\/em> + 9d<\/em>
It is given that
a<\/em>17<\/sub> \u2212 a<\/em>10<\/sub> = 7
(a<\/em> + 16d<\/em>) \u2212 (a<\/em> + 9d<\/em>) = 7
7d<\/em> = 7
d<\/em> = 1
Therefore, the common difference is 1.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

11. Which term of the A.P. 3, 15, 27, 39, \u2026 will be 132 more than its 54th<\/sup> term?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given A.P. is 3, 15, 27, 39, \u2026
a <\/em>= 3
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 15 \u2212 3 = 12
a<\/em>54<\/sub> = a<\/em> + (54 \u2212 1) d<\/em>
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let n<\/em>th<\/sup> term be 771.
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
771 = 3 + (n<\/em> \u2212 1) 12
768 = (n<\/em> \u2212 1) 12
(n<\/em> \u2212 1) = 64
n<\/em> = 65
Therefore, 65th<\/sup> term was 132 more than 54th<\/sup> term.<\/p>\n\n\n\n

Or<\/strong>
Let n<\/em>th<\/sup> term be 132 more than 54th<\/sup> term.
n<\/em> = 54 + 132\/2
= 54 + 11 = 65th<\/sup> term<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

12. Two APs have the same common difference. The difference between their 100th<\/sup> term is 100, what is the difference between their 1000th<\/sup> terms?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Let the first term of these A.P.s be a<\/em>1<\/sub> and a<\/em>2<\/sub> respectively and the common difference of these A.P.s be d<\/em>.
For first A.P.,
a<\/em>100<\/sub> = a<\/em>1<\/sub> + (100 \u2212 1) d<\/em>
= a<\/em>1<\/sub> + 99d
a<\/em>1000<\/sub> = a<\/em>1<\/sub> + (1000 \u2212 1) d<\/em>
a<\/em>1000<\/sub> = a<\/em>1<\/sub> + 999d<\/em>
For second A.P.,
a<\/em>100<\/sub> = a<\/em>2<\/sub> + (100 \u2212 1) d<\/em>
= a<\/em>2<\/sub> + 99d<\/em>
a<\/em>1000<\/sub> = a<\/em>2<\/sub> + (1000 \u2212 1) d<\/em>
= a<\/em>2<\/sub> + 999d<\/em>
Given that, difference between
100th<\/sup> term of these A.P.s = 100
Therefore, (a<\/em>1<\/sub> + 99d<\/em>) \u2212 (a<\/em>2<\/sub> + 99d<\/em>) = 100
a<\/em>1<\/sub> \u2212 a<\/em>2<\/sub> = 100 … (i)<\/strong>
Difference between 1000th<\/sup> terms of these A.P.s
(a<\/em>1<\/sub> + 999d<\/em>) \u2212 (a<\/em>2<\/sub> + 999d<\/em>) = a<\/em>1<\/sub> \u2212 a<\/em>2<\/sub>
From equation (i)<\/strong>,
This difference, a<\/em>1<\/sub> \u2212 a<\/em>2 <\/sub>= 100
Hence, the difference between 1000th<\/sup> terms of these A.P. will be 100.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

13. How many three digit numbers are divisible by 7?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, \u2026
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 \u2212 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, \u2026, 994
Let 994 be the n<\/em>th term of this A.P.
a<\/em> = 105
d<\/em> = 7
a<\/em>n<\/em><\/sub> = 994
n<\/em> = ?
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
994 = 105 + (n<\/em> \u2212 1) 7
889 = (n<\/em> \u2212 1) 7
(n <\/em>\u2212 1) = 127
n<\/em> = 128
Therefore, 128 three-digit numbers are divisible by 7.<\/p>\n\n\n\n

Or<\/strong><\/p>\n\n\n\n

Three digit numbers which are divisible by 7 are 105, 112, 119, …. 994 .
These numbers form an AP with a<\/em> = 105 and d<\/em> = 7.
Let number of three-digit numbers divisible by 7 be n<\/em>, a<\/em>n<\/sub> = 994
\u21d2 a<\/em> + (n<\/em> – 1) d<\/em> = 994
\u21d2 105 + (n<\/em> – 1) \u00d7 7 = 994
\u21d27(n<\/em> – 1) = 889
\u21d2 n<\/em> – 1 = 127
\u21d2 n<\/em> = 128<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

14. How many multiples of 4 lie between 10 and 250?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, \u2026
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 \u2212 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, \u2026, 248
Let 248 be the n<\/em>th<\/sup> term of this A.P.
a<\/em> = 12
d<\/em> = 4
a<\/em>n<\/sub> = 248
a<\/em>n<\/sub> = a<\/em> + (n<\/em> – 1) d<\/em>
248 = 12 + (n<\/em> – 1) \u00d7 4
236\/4 = n<\/em> – 1
59  = n<\/em> – 1
n<\/em> = 60
Therefore, there are 60 multiples of 4 between 10 and 250.<\/p>\n\n\n\n

Or<\/strong><\/p>\n\n\n\n

Multiples of 4 lies between 10 and 250 are 12, 16, 20, …., 248.
These numbers form an AP with a<\/em> = 12 and d<\/em> = 4.
Let number of three-digit numbers divisible by 4 be n<\/em>, a<\/em>n<\/sub> = 248
\u21d2 a<\/em> + (n<\/em> – 1) d<\/em> = 248
\u21d2 12 + (n<\/em> – 1) \u00d7 4 = 248
\u21d24(n<\/em> – 1) = 248
\u21d2 n<\/em> – 1 = 59
\u21d2 n<\/em> = 60<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

15. For what value of n<\/em>, are the n<\/em>th<\/sup> terms of two APs 63, 65, 67, and 3, 10, 17, \u2026 equal?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
63, 65, 67, \u2026
a<\/em> = 63
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 65 \u2212 63 = 2
n<\/em>th<\/sup> term of this A.P. = a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
a<\/em>n<\/em><\/sub>= 63 + (n<\/em> \u2212 1) 2 = 63 + 2n<\/em> \u2212 2
a<\/em>n<\/em><\/sub> = 61 + 2n<\/em> … (i)<\/strong>
3, 10, 17, \u2026
a<\/em> = 3
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 10 \u2212 3 = 7
n<\/em>th<\/sup> term of this A.P. = 3 + (n<\/em> \u2212 1) 7
a<\/em>n<\/em><\/sub>= 3 + 7n<\/em> \u2212 7
a<\/em>n<\/em><\/sub> = 7n<\/em> \u2212 4 … (ii)<\/strong>
It is given that, n<\/em>th<\/sup> term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n<\/em> = 7n<\/em> \u2212 4
61 + 4 = 5n<\/em>
5n<\/em> = 65
n<\/em> = 13
Therefore, 13th<\/sup> terms of both these A.P.s are equal to each other.
16. Determine the A.P. whose third term is 16 and the 7th<\/sup> term exceeds the 5th<\/sup> term by 12.<\/p>\n\n\n\n

a<\/em> + (3 \u2212 1) d<\/em> = 16
a<\/em> + 2d<\/em> = 16 … (i)<\/strong>
a<\/em>7<\/sub> \u2212 a<\/em>5<\/sub> = 12
[a<\/em>+ (7 \u2212 1) d<\/em>] \u2212 [a <\/em>+ (5 \u2212 1) d<\/em>]= 12
(a<\/em> + 6d<\/em>) \u2212 (a<\/em> + 4d<\/em>) = 12
2d<\/em> = 12
d<\/em> = 6
From equation (i)<\/strong>, we get,
a<\/em> + 2 (6) = 16
a<\/em> + 12 = 16
a<\/em> = 4
Therefore, A.P. will be
4, 10, 16, 22, \u2026<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

17. Find the 20th<\/sup> term from the last term of the A.P. 3, 8, 13, \u2026, 253.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given A.P. is
3, 8, 13, \u2026, 253
Common difference for this A.P. is 5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, \u2026, 13, 8, 5
For this A.P.,
a<\/em> = 253
d<\/em> = 248 \u2212 253 = \u22125
n <\/em>= 20
a<\/em>20<\/sub> = a<\/em> + (20 \u2212 1) d<\/em>
a<\/em>20<\/sub> = 253 + (19) (\u22125)
a<\/em>20<\/sub> = 253 \u2212 95
a<\/em> = 158
Therefore, 20th<\/sup> term from the last term is 158.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/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 44. Find the first three terms of the A.P.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
We know that,
a<\/em>n<\/em><\/sub> = a<\/em> + (n <\/em>\u2212 1) d<\/em>
a<\/em>4<\/sub> = a<\/em> + (4 \u2212 1) d<\/em>
a<\/em>4<\/sub> = a<\/em> + 3d<\/em>
Similarly,
a<\/em>8<\/sub> = a<\/em> + 7d<\/em>
a<\/em>6<\/sub> = a<\/em> + 5d<\/em>
a<\/em>10<\/sub> = a<\/em> + 9d<\/em>
Given that, a<\/em>4<\/sub> + a<\/em>8<\/sub> = 24
a<\/em> + 3d<\/em> + a <\/em>+ 7d<\/em> = 24
2a<\/em> + 10d<\/em> = 24
a<\/em> + 5d<\/em> = 12 … (i)<\/strong>
a<\/em>6<\/sub> + a<\/em>10<\/sub> = 44
a<\/em> + 5d<\/em> + a<\/em> + 9d<\/em> = 44
2a<\/em> + 14d<\/em> = 44
a<\/em> + 7d<\/em> = 22 … (ii)<\/strong>
On subtracting equation (i)<\/strong> from (ii)<\/strong>, we get,
2d<\/em> = 22 \u2212 12
2d<\/em> = 10
d<\/em> = 5
From equation (i)<\/strong>, we get
a<\/em> + 5d<\/em> = 12
a<\/em> + 5 (5) = 12
a<\/em> + 25 = 12
a<\/em> = \u221213
a<\/em>2<\/sub> = a<\/em> + d<\/em> = \u2212 13 + 5 = \u22128
a<\/em>3<\/sub> = a<\/em>2<\/sub> + d<\/em> = \u2212 8 + 5 = \u22123
Therefore, the first three terms of this A.P. are \u221213, \u22128, and \u22123.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, \u2026
Here, a<\/em> = 5000
d<\/em> = 200
Let after n<\/em>th<\/sup> year, his salary be Rs 7000.
Therefore, a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
7000 = 5000 + (n<\/em> \u2212 1) 200
200(n<\/em> \u2212 1) = 2000
(n<\/em> \u2212 1) = 10
n<\/em> = 11
Therefore, in 11th year, his salary will be Rs 7000.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n<\/em>th<\/sup> week, her week, her weekly savings become Rs 20.75, find n.<\/em><\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given that,
a<\/em> = 5
d <\/em>= 1.75
a<\/em>n <\/em><\/sub>= 20.75
n<\/em> = ?
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
20.75 = 5 + (n<\/em> – 1) \u00d7 1.75
15.75 = (n<\/em> – 1) \u00d7 1.75
(n<\/em> – 1) = 15.75\/1.75 = 1575\/175
= 63\/7 = 9
n<\/em> – 1 = 9
n<\/em> = 10
Hence, n<\/em> is 10.<\/p>\n\n\n\n

1. Find the sum of the following APs.<\/strong>
(i) 2, 7, 12 ,\u2026., to 10 terms.
(ii) \u2212 37, \u2212 33, \u2212 29 ,\u2026, to 12 terms
(iii) 0.6, 1.7, 2.8 ,\u2026\u2026.., to 100 terms
(iv) 1\/15, 1\/12, 1\/10, …… , to 11 terms<\/p>\n\n\n\n

Answer<\/strong>
(i) 2, 7, 12 ,\u2026, to 10 terms
For this A.P.,
a<\/em> = 2
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 7 \u2212 2 = 5
n<\/em> = 10
We know that,
Sn<\/sub><\/em> = n<\/em>\/2 [2a + (n<\/em> – 1) d<\/em>]
S10<\/sub><\/em> = 10\/2 [2(2) + (10 – 1) \u00d7 5]
= 5[4 + (9) \u00d7 (5)]
= 5 \u00d7 49 = 245<\/p>\n\n\n\n

(ii) \u221237, \u221233, \u221229 ,\u2026, to 12 terms
For this A.P.,
a<\/em> = \u221237
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = (\u221233) \u2212 (\u221237)
= \u2212 33 + 37 = 4
n<\/em> = 12
We know that,
Sn<\/sub><\/em> = n<\/em>\/2 [2a + (n<\/em> – 1) d<\/em>]
S12<\/sub><\/em> = 12\/2 [2(-37) + (12 – 1) \u00d7 4]
= 6[-74 + 11 \u00d7 4]
= 6[-74 + 44]
= 6(-30) = -180<\/p>\n\n\n\n

(iii) 0.6, 1.7, 2.8 ,\u2026, to 100 terms
For this A.P.,
a<\/em> = 0.6
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 1.7 \u2212 0.6 = 1.1
n<\/em> = 100
We know that,
Sn<\/sub><\/em> = n<\/em>\/2 [2a + (n<\/em> – 1) d<\/em>]
S12<\/sub><\/em> = 50\/2 [1.2 + (99) \u00d7 1.1]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505<\/p>\n\n\n\n

(iv) 1\/15, 1\/12, 1\/10, …… , to 11 terms
For this A.P.,
<\/p>\n\n\n\n

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

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

2. Find the sums given below
<\/strong>(i) 7 +<\/a> + 14 + ……………… +84
(ii)+ 14 + \u2026\u2026\u2026\u2026 + 84
(ii) 34 + 32 + 30 + \u2026\u2026\u2026.. + 10
(iii) \u2212 5 + (\u2212 8) + (\u2212 11) + \u2026\u2026\u2026\u2026 + (\u2212 230)<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) For this A.P.,
a<\/em> = 7
l<\/em> = 84
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> =<\/p>\n\n\n\n

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

 – 7 = 21\/2 – 7 = 7\/2
Let 84 be the n<\/em>th <\/sup>term of this A.P.
l<\/em> = a<\/em> (n<\/em> – 1)d<\/em>
84 = 7 + (n<\/em> – 1) \u00d7 7\/2
77 = (n<\/em> – 1) \u00d7 7\/2
22 = n<\/em> \u2212 1
n<\/em> = 23
We know that,
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + l<\/em>)
Sn<\/sub><\/em> = 23\/2 (7 + 84)
= (23\u00d791\/2) = 2093\/2
= <\/p>\n\n\n\n

(ii) 34 + 32 + 30 + \u2026\u2026\u2026.. + 10
For this A.P.,
a<\/em> = 34
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 32 \u2212 34 = \u22122
l<\/em> = 10
Let 10 be the n<\/em>th<\/sup> term of this A.P.
l<\/em> = a<\/em> + (n <\/em>\u2212 1) d<\/em>
10 = 34 + (n<\/em> \u2212 1) (\u22122)
\u221224 = (n <\/em>\u2212 1) (\u22122)
12 = n<\/em> \u2212 1
n<\/em> = 13
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + l<\/em>)
= 13\/2 (34 + 10)
= (13\u00d744\/2) = 13 \u00d7 22
= 286<\/p>\n\n\n\n

(iii) (\u22125) + (\u22128) + (\u221211) + \u2026\u2026\u2026\u2026 + (\u2212230) For this A.P.,
a <\/em>= \u22125
l<\/em> = \u2212230
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = (\u22128) \u2212 (\u22125)
= \u2212 8 + 5 = \u22123
Let \u2212230 be the n<\/em>th<\/sup> term of this A.P.
l<\/em> = a<\/em> + (n <\/em>\u2212 1)d<\/em>
\u2212230 = \u2212 5 + (n<\/em> \u2212 1) (\u22123)
\u2212225 = (n<\/em> \u2212 1) (\u22123)
(n<\/em> \u2212 1) = 75
n<\/em> = 76
And,
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + l<\/em>)
= 76\/2 [(-5) + (-230)]
= 38(-235)
= -8930<\/p>\n\n\n\n

3. In an AP
<\/strong>(i) Given a<\/em> = 5, d<\/em> = 3, a<\/em>n<\/em><\/sub> = 50, find n<\/em> and S<\/em>n<\/em><\/sub>.
(ii) Given a<\/em> = 7, a<\/em>13<\/sub> = 35, find d<\/em> and S<\/em>13<\/sub>.
(iii) Given a<\/em>12<\/sub> = 37, d<\/em> = 3, find a<\/em> and S<\/em>12<\/sub>.
(iv) Given a<\/em>3<\/sub> = 15, S<\/em>10<\/sub> = 125, find d<\/em> and a<\/em>10<\/sub>.
(v) Given d<\/em> = 5, S<\/em>9<\/sub> = 75, find a<\/em> and a<\/em>9<\/sub>.
(vi) Given a<\/em> = 2, d<\/em> = 8, S<\/em>n<\/em><\/sub> = 90, find n<\/em> and a<\/em>n<\/em><\/sub>.
(vii) Given a<\/em> = 8, a<\/em>n<\/em><\/sub> = 62, S<\/em>n<\/em><\/sub> = 210, find n<\/em> and d<\/em>.
(viii) Given a<\/em>n<\/em><\/sub> = 4, d<\/em> = 2, S<\/em>n<\/em><\/sub> = \u2212 14, find n<\/em> and a<\/em>.
(ix) Given a<\/em> = 3, n<\/em> = 8, S<\/em> = 192, find d<\/em>.
(x) Given l<\/em> = 28, S<\/em> = 144 and there are total 9 terms. Find a<\/em>.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Given that, a<\/em> = 5, d<\/em> = 3, a<\/em>n<\/em><\/sub> = 50
As a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>,
\u21d2 50 = 5 + (n<\/em> – 1) \u00d7 3
\u21d2 3(n<\/em> – 1) = 45
\u21d2 n<\/em> – 1 = 15
\u21d2 n<\/em> = 16
Now, Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + a<\/em>n<\/em><\/sub>)
Sn<\/sub><\/em> = 16\/2 (5 + 50) = 440<\/p>\n\n\n\n

(ii) Given that, a<\/em> = 7, a<\/em>13<\/sub> = 35
As a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>, \u21d2 35 = 7 + (13 – 1)d<\/em>
\u21d2 12d<\/em> = 28
\u21d2 d<\/em> = 28\/12 = 2.33
Now, Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + a<\/em>n<\/em><\/sub>)
S13<\/sub><\/em> = 13\/2 (7 + 35) = 273<\/p>\n\n\n\n

(iii)Given that, a<\/em>12<\/sub> = 37, d<\/em> = 3 As a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>,
\u21d2 a<\/em>12<\/sub> = a<\/em> + (12 \u2212 1)3
\u21d2 37 = a<\/em> + 33
\u21d2 a<\/em> = 4
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + a<\/em>n<\/em><\/sub>)
Sn<\/sub><\/em> = 12<\/em>\/2 (4 + 37)
= 246<\/p>\n\n\n\n

(iv) Given that, a<\/em>3<\/sub> = 15, S<\/em>10<\/sub> = 125
As a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>,
a<\/em>3<\/sub> = a<\/em> + (3 \u2212 1)d<\/em>
15 = a<\/em> + 2d<\/em> … (i)<\/strong>
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S10<\/sub><\/em> = 10\/2 [2a<\/em> + (10 – 1)d<\/em>]
125 = 5(2a<\/em> + 9d<\/em>)
25 = 2a<\/em> + 9d <\/em>… (ii)<\/strong>
On multiplying equation (i)<\/strong> by (ii)<\/strong>, we get
30 = 2a<\/em> + 4d<\/em> … (iii)<\/strong>
On subtracting equation (iii)<\/strong> from (ii)<\/strong>, we get
\u22125 = 5d<\/em>
d<\/em> = \u22121
From equation (i)<\/strong>,
15 = a<\/em> + 2(\u22121)
15 = a<\/em> \u2212 2
a<\/em> = 17
a<\/em>10<\/sub> = a<\/em> + (10 \u2212 1)d<\/em>
a<\/em>10<\/sub> = 17 + (9) (\u22121)
a<\/em>10<\/sub> = 17 \u2212 9 = 8<\/p>\n\n\n\n

(v) Given that, d<\/em> = 5, S<\/em>9<\/sub> = 75
As Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S9<\/sub><\/em> = 9\/2 [2a<\/em> + (9 – 1)5<\/em>]
25 = 3(a<\/em> + 20)
25 = 3a<\/em> + 60
3a<\/em> = 25 \u2212 60
a<\/em> = -35\/3
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>
a<\/em>9<\/sub> = a<\/em> + (9 \u2212 1) (5)
= -35\/3 + 8(5)
= -35\/3 + 40
= (35+120\/3) = 85\/3<\/p>\n\n\n\n

(vi) Given that, a<\/em> = 2, d<\/em> = 8, S<\/em>n<\/em><\/sub> = 90
As Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
90 = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
\u21d2 180 = n<\/em>(4 + 8n<\/em> – 8) = n<\/em>(8n<\/em> – 4) = 8n<\/em>2<\/sup> – 4n<\/em>
\u21d2 8n<\/em>2<\/sup> – 4n –<\/em> 180 = 0
\u21d2 2n<\/em>2<\/sup> – n<\/em> – 45 = 0
\u21d2 2n<\/em>2<\/sup> – 10n<\/em> + 9n<\/em> – 45 = 0
\u21d2 2n<\/em>(n<\/em> -5) + 9(n<\/em> – 5) = 0
\u21d2 (2n<\/em> – 9)(2n<\/em> + 9) = 0
So, n<\/em> = 5 (as it is positive integer)
\u2234 a<\/em>5<\/em> <\/sub>= 8 + 5 \u00d7 4 = 34<\/p>\n\n\n\n

(vii) Given that, a<\/em> = 8, a<\/em>n<\/em><\/sub> = 62, S<\/em>n<\/em><\/sub> = 210
As Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + a<\/em>n<\/em><\/sub>)
210 = n<\/em>\/2 (8 + 62)
\u21d2 35n<\/em> = 210
\u21d2 n<\/em> = 210\/35 = 6
Now, 62 = 8 + 5d<\/em>
\u21d2 5d<\/em> = 62 – 8 = 54
\u21d2 d<\/em> = 54\/5 = 10.8<\/p>\n\n\n\n

(viii) Given that, a<\/em>n<\/em><\/sub> = 4, d<\/em> = 2, S<\/em>n<\/em><\/sub> = \u221214
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>
4 = a<\/em> + (n <\/em>\u2212 1)2
4 = a<\/em> + 2n<\/em> \u2212 2
a<\/em> + 2n<\/em> = 6
a <\/em>= 6 \u2212 2n<\/em> … (i)<\/strong>
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + a<\/em>n<\/em><\/sub>)
-14 = n<\/em>\/2 (a<\/em> + 4<\/em>)
\u221228 = n <\/em>(a<\/em> + 4)
\u221228 = n <\/em>(6 \u2212 2n<\/em> + 4) {From equation (i)<\/strong>}
\u221228 = n <\/em>(\u2212 2n<\/em> + 10)
\u221228 = \u2212 2n<\/em>2<\/sup> + 10n<\/em>
2n<\/em>2<\/sup> \u2212 10n<\/em> \u2212 28 = 0
n<\/em>2<\/sup> \u2212 5n <\/em>\u221214 = 0
n<\/em>2<\/sup> \u2212 7n + <\/em>2n<\/em> \u2212 14 = 0
n <\/em>(n<\/em> \u2212 7) + 2(n<\/em> \u2212 7) = 0
(n<\/em> \u2212 7) (n<\/em> + 2) = 0
Either n<\/em> \u2212 7 = 0 or n<\/em> + 2 = 0
n<\/em> = 7 or n<\/em> = \u22122
However, n<\/em> can neither be negative nor fractional.
Therefore, n<\/em> = 7
From equation (i)<\/strong>, we get
a<\/em> = 6 \u2212 2n<\/em>
a<\/em> = 6 \u2212 2(7)
= 6 \u2212 14
= \u22128<\/p>\n\n\n\n

(ix) Given that, a<\/em> = 3, n<\/em> = 8, S<\/em> = 192
As Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
192 = 8\/2 [2 \u00d7 3 + (8 – 1)d<\/em>]
192 = 4 [6 + 7d<\/em>]
48 = 6 + 7d<\/em>
42 = 7d<\/em>
d = <\/em>6<\/p>\n\n\n\n

(x) Given that, l<\/em> = 28, S<\/em> = 144 and there are total of 9 terms.
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + l<\/em>)
144 = 9\/2 (a<\/em> + 28)
(16) \u00d7 (2) = a<\/em> + 28
32 = a<\/em> + 28
a<\/em> = 4<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

4. How many terms of the AP. 9, 17, 25 \u2026 must be taken to give a sum of 636?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Let there be n<\/em> terms of this A.P.
For this A.P., a<\/em> = 9
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 17 \u2212 9 = 8
As Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
636 = n<\/em>\/2 [2 \u00d7 a<\/em> + (8 – 1) \u00d7 8]
636 = n<\/em>\/2 [18 + (n<\/em>– 1) \u00d7 8]
636 = n <\/em>[9 + 4n<\/em> \u2212 4]
636 = n <\/em>(4n<\/em> + 5)
4n<\/em>2<\/sup> + 5n<\/em> \u2212 636 = 0
4n<\/em>2<\/sup> + 53n<\/em> \u2212 48n<\/em> \u2212 636 = 0
n <\/em>(4n<\/em> + 53) \u2212 12 (4n<\/em> + 53) = 0
(4n<\/em> + 53) (n<\/em> \u2212 12) = 0
Either 4n <\/em>+ 53 = 0 or n<\/em> \u2212 12 = 0
n<\/em> = (-53\/4) or n<\/em> = 12
n <\/em>cannot be (-53\/4). As the number of terms can neither be negative nor fractional, therefore, n<\/em> = 12 only.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

5. The first term of an AP 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

Answer<\/strong>
Given that,
a<\/em> = 5
l<\/em> = 45
S<\/em>n<\/em><\/sub> = 400
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + l<\/em>)
400 = n<\/em>\/2 (5 + 45)
400 = n<\/em>\/2 (50)
n<\/em> = 16
l = a + <\/em>(n<\/em> \u2212 1) d<\/em>
45 = 5 + (16 \u2212 1) d<\/em>
40 = 15d<\/em>
d<\/em> = 40\/15 = 8\/3<\/p>\n\n\n\n

6. The first and the last term of an AP 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

Answer<\/strong>
Given that,
a<\/em> = 17
l<\/em> = 350
d<\/em> = 9
Let there be n<\/em> terms in the A.P.
l = a + <\/em>(n<\/em> \u2212 1) d<\/em>
350 = 17 + (n<\/em> \u2212 1)9
333 = (n<\/em> \u2212 1)9
(n<\/em> \u2212 1) = 37
n<\/em> = 38
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + l<\/em>)
S38<\/sub><\/em> = 13\/2 (17 + 350)
= 19 \u00d7 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

7. Find the sum of first 22 terms of an AP in which d<\/em> = 7 and 22nd<\/sup> term is 149.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
d<\/em> = 7
a<\/em>22<\/sub> = 149
S<\/em>22<\/sub> = ?
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>
a<\/em>22<\/sub> = a<\/em> + (22 \u2212 1)d<\/em>
149 = a<\/em> + 21 \u00d7 7
149 = a<\/em> + 147
a<\/em> = 2
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + a<\/em>n<\/em><\/sub>)
= 22\/2 (2 + 149)
= 11 \u00d7 151
= 1661<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5<\/p>\n\n\n\n

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given that,
a<\/em>2<\/sub> = 14
a<\/em>3<\/sub> = 18
d<\/em> = a<\/em>3<\/sub> \u2212 a<\/em>2<\/sub> = 18 \u2212 14 = 4
a<\/em>2<\/sub> = a<\/em> + d<\/em>
14 = a<\/em> + 4
a<\/em> = 10
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S51<\/sub><\/em> = 51\/2 [2 \u00d7 10 + (51 – 1) \u00d7 4]
= 51\/2 [2 + (20) \u00d7 4]
= 51\u00d7220\/2
= 51 \u00d7 110
= 5610<\/p>\n\n\n\n

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n<\/em> terms.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given that,
S<\/em>7<\/sub> = 49
S17<\/sub> = 289
S7<\/sub><\/em>  = 7<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S7<\/sub><\/em> = 7\/2 [2a<\/em> + (7 – 1)d<\/em>]
49 = 7\/2 [2a<\/em> + 16d<\/em>]
7 = (a<\/em> + 3d<\/em>)
a<\/em> + 3d<\/em> = 7 … (i)<\/strong>
Similarly,
S17<\/sub><\/em> = 17\/2 [2a<\/em> + (17 – 1)d<\/em>]
289 = 17\/2 (2a<\/em> + 16d<\/em>)
17 = (a<\/em> + 8d<\/em>)
a<\/em> + 8d<\/em> = 17 … (ii)<\/strong>
Subtracting equation (i)<\/strong> from equation (ii)<\/strong>,
5d<\/em> = 10
d<\/em> = 2
From equation (i)<\/strong>,
a<\/em> + 3(2) = 7
a + <\/em>6 = 7
a = <\/em>1
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
= n<\/em>\/2 [2(1) + (n<\/em> – 1) \u00d7 2]
= n<\/em>\/2 (2 + 2n<\/em> – 2)
= n<\/em>\/2 (2n<\/em>)
= n<\/em>2<\/sup>
<\/p>\n\n\n\n

10. Show that a<\/em>1<\/sub>, a<\/em>2 <\/sub>\u2026 , a<\/em>n<\/em><\/sub> , \u2026 form an AP where a<\/em>n<\/em><\/sub> is defined as below<\/strong>
(i) a<\/em>n<\/em><\/sub> = 3 + 4n<\/em>
(ii) a<\/em>n<\/em><\/sub> = 9 \u2212 5n<\/em>
Also find the sum of the first 15 terms in each case.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) a<\/em>n<\/em><\/sub> = 3 + 4n<\/em>
a<\/em>1<\/sub> = 3 + 4(1) = 7
a<\/em>2<\/sub> = 3 + 4(2) = 3 + 8 = 11
a<\/em>3<\/sub> = 3 + 4(3) = 3 + 12 = 15
a<\/em>4<\/sub> = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 11 \u2212 7 = 4
a<\/em>3<\/sub> \u2212 a<\/em>2<\/sub> = 15 \u2212 11 = 4
a<\/em>4<\/sub> \u2212 a<\/em>3<\/sub> = 19 \u2212 15 = 4
i.e., a<\/em>k<\/em><\/sub> + 1<\/sub> \u2212 a<\/em>k<\/em><\/sub> is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S15 <\/sub><\/em>= 15\/2 [2(7) + (15 – 1) \u00d7 4]
= 15\/2 [(14) + 56]
= 15\/2 (70)
= 15 \u00d7 35
= 525<\/p>\n\n\n\n

(ii) a<\/em>n<\/em><\/sub> = 9 \u2212 5n<\/em>
a<\/em>1<\/sub> = 9 \u2212 5 \u00d7 1 = 9 \u2212 5 = 4
a<\/em>2<\/sub> = 9 \u2212 5 \u00d7 2 = 9 \u2212 10 = \u22121
a<\/em>3<\/sub> = 9 \u2212 5 \u00d7 3 = 9 \u2212 15 = \u22126
a<\/em>4<\/sub> = 9 \u2212 5 \u00d7 4 = 9 \u2212 20 = \u221211
It can be observed that
a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = \u2212 1 \u2212 4 = \u22125
a<\/em>3<\/sub> \u2212 a<\/em>2<\/sub> = \u2212 6 \u2212 (\u22121) = \u22125
a<\/em>4<\/sub> \u2212 a<\/em>3<\/sub> = \u2212 11 \u2212 (\u22126) = \u22125
i.e., a<\/em>k<\/em><\/sub> + 1<\/sub> \u2212 a<\/em>k<\/em><\/sub> is same every time. Therefore, this is an A.P. with common difference as \u22125 and first term as 4.
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S15 <\/sub><\/em>= 15\/2 [2(4) + (15 – 1) (-5)]
= 15\/2 [8 + 14(-5)]
= 15\/2 (8 – 70)
= 15\/2 (-62)
= 15(-31)
= -465<\/p>\n\n\n\n

11. If the sum of the first n<\/em> terms of an AP is 4n<\/em> \u2212 n<\/em>2<\/sup>, what is the first term (that is S<\/em>1<\/sub>)? What is the sum of first two terms? What is the second term? Similarly find the 3rd<\/sup>, the10th<\/sup> and the n<\/em>th<\/sup> terms.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Given that,
S<\/em>n<\/em><\/sub> = 4n<\/em> \u2212 n<\/em>2<\/sup>
First term, a<\/em> = S<\/em>1<\/sub> = 4(1) \u2212 (1)2<\/sup> = 4 \u2212 1 = 3
Sum of first two terms = S<\/em>2<\/sub>
= 4(2) \u2212 (2)2<\/sup> = 8 \u2212 4 = 4
Second term, a<\/em>2<\/sub> = S<\/em>2<\/sub> \u2212 S<\/em>1<\/sub> = 4 \u2212 3 = 1
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em> = 1 \u2212 3 = \u22122
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>
= 3 + (n<\/em> \u2212 1) (\u22122)
= 3 \u2212 2n<\/em> + 2
= 5 \u2212 2n<\/em>
Therefore, a<\/em>3<\/sub> = 5 \u2212 2(3) = 5 \u2212 6 = \u22121
a<\/em>10<\/sub> = 5 \u2212 2(10) = 5 \u2212 20 = \u221215
Hence, the sum of first two terms is 4. The second term is 1. 3rd<\/sup>, 10th<\/sup>, and n<\/em>th<\/sup> terms are \u22121, \u221215, and 5 \u2212 2n<\/em> respectively.<\/p>\n\n\n\n

12. Find the sum of first 40 positive integers divisible by 6.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
The positive integers that are divisible by 6 are
6, 12, 18, 24 \u2026
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a<\/em> = 6
d<\/em> = 6
S<\/em>40<\/sub> = <\/em>?
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S40 <\/sub><\/em>= 40\/2 [2(6) + (40 – 1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 \u00d7 246
= 4920<\/p>\n\n\n\n

13. Find the sum of first 15 multiples of 8.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
The multiples of 8 are
8, 16, 24, 32\u2026
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a<\/em> = 8
d<\/em> = 8
S<\/em>15<\/sub> = ?
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S15<\/sub> <\/em>= 15\/2 [2(8) + (15 – 1)8]
= 15\/2[6 + (14) (8)]
= 15\/2[16 + 112]
= 15(128)\/2
= 15 \u00d7 64
= 960<\/p>\n\n\n\n

14. Find the sum of the odd numbers between 0 and 50.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 \u2026 49
Therefore, it can be observed that these odd numbers are in an A.P.
a<\/em> = 1
d<\/em> = 2
l<\/em> = 49
l<\/em> = a<\/em> + (n<\/em> \u2212 1) d<\/em>
49 = 1 + (n<\/em> \u2212 1)2
48 = 2(n<\/em> \u2212 1)
n<\/em> \u2212 1 = 24
n<\/em> = 25
Sn<\/sub><\/em> = n<\/em>\/2 (a<\/em> + l<\/em>)
S25<\/sub><\/em> = 25\/2 (1 + 49)
= 25(50)\/2
=(25)(25)
= 625<\/p>\n\n\n\n

15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a<\/em> = 200
d<\/em> = 50
Penalty that has to be paid if he has delayed the work by 30 days = S<\/em>30<\/sub>
= 30\/2 [2(200) + (30 – 1) 50]

= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.<\/p>\n\n\n\n

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.<\/strong><\/p>\n\n\n\n

Answer<\/strong>
Let the cost of 1st<\/sup> prize be P<\/em>.
Cost of 2nd<\/sup> prize = P<\/em> \u2212 20
And cost of 3rd<\/sup> prize = P<\/em> \u2212 40
It can be observed that the cost of these prizes are in an A.P. having common difference as \u221220 and first term as P<\/em>.
a<\/em> = P<\/em>
d<\/em> = \u221220
Given that, S<\/em>7<\/sub> = 700
7\/2 [2a<\/em> + (7 – 1)d<\/em>] = 700

a<\/em> + 3(\u221220) = 100
a<\/em> \u2212 60 = 100
a<\/em> = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.<\/p>\n\n\n\n

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5\u2026\u2026\u2026\u2026\u2026\u2026..12
First term, a<\/em> = 1
Common difference, d<\/em> = 2 \u2212 1 = 1
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S12<\/sub> <\/em>= 12\/2 [2(1) + (12 – 1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 \u00d7 78 = 234
Therefore, 234 trees will be planted by the students.<\/p>\n\n\n\n

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, \u2026\u2026\u2026 as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \u03c0 = 22\/7)<\/strong><\/p>\n\n\n\n

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

Answer<\/strong>
perimeter of semi-circle = \u03c0r<\/em>
P<\/em>1<\/sub> = \u03c0(0.5) = \u03c0\/2 cm
P<\/em>2<\/sub> = \u03c0(1) = \u03c0 cm
P<\/em>3<\/sub> = \u03c0(1.5) = 3\u03c0\/2 cm
P<\/em>1<\/sub>, P<\/em>2<\/sub>, P<\/em>3<\/sub> are the lengths of the semi-circles
\u03c0\/2, \u03c0, 3\u03c0\/2, 2\u03c0, ….
P<\/em>1= \u03c0\/2 cm
P<\/em>2<\/sub> = \u03c0 cm
d<\/em> = P2- P<\/em>1 = \u03c0 – \u03c0\/2 = \u03c0\/2
First term = P<\/em>1 = a<\/em> = \u03c0\/2 cm
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
Therefor, Sum of the length of 13 consecutive circles
S13<\/sub><\/em> = 13<\/em>\/2 [2(\u03c0\/2) + (13 – 1)\u03c0\/2]
=  13<\/em>\/2 [\u03c0 + 6\u03c0]
=13<\/em>\/2 (7\u03c0)  = 13<\/em>\/2 \u00d7 7 \u00d7 22<\/em>\/7
= 143 cm<\/p>\n\n\n\n

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?<\/strong><\/p>\n\n\n\n

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

Answer<\/strong>
It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18\u2026
For this A.P.,
a<\/em> = 20
d<\/em> = a<\/em>2<\/sub> \u2212 a<\/em>1<\/sub> = 19 \u2212 20 = \u22121
Let a total of 200 logs be placed in n<\/em> rows.
S<\/em>n<\/em><\/sub> = 200
Sn<\/sub><\/em> = n<\/em>\/2 [2a<\/em> + (n<\/em> – 1)d<\/em>]
S12<\/sub> <\/em>= 12\/2 [2(20) + (n<\/em> – 1)(-1)]
400 = n<\/em> (40 \u2212 n<\/em> + 1)
400 = n <\/em>(41 \u2212 n<\/em>)
400 = 41n<\/em> \u2212 n<\/em>2<\/sup>
n<\/em>2<\/sup> \u2212 41n <\/em>+ 400 = 0
n<\/em>2<\/sup> \u2212 16n<\/em> \u2212 25n<\/em> + 400 = 0
n <\/em>(n<\/em> \u2212 16) \u221225 (n<\/em> \u2212 16) = 0
(n <\/em>\u2212 16) (n<\/em> \u2212 25) = 0
Either (n<\/em> \u2212 16) = 0 or n<\/em> \u2212 25 = 0
n<\/em> = 16 or n<\/em> = 25
a<\/em>n<\/em><\/sub> = a<\/em> + (n<\/em> \u2212 1)d<\/em>
a<\/em>16<\/sub> = 20 + (16 \u2212 1) (\u22121)
a<\/em>16<\/sub> = 20 \u2212 15
a<\/em>16<\/sub> = 5
Similarly,
a<\/em>25<\/sub> = 20 + (25 \u2212 1) (\u22121)
a<\/em>25<\/sub> = 20 \u2212 24
= \u2212<\/em>4
Clearly, the number of logs in 16th<\/sup> row is 5. However, the number of logs in 25th<\/sup> row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th<\/sup> row is 5.<\/p>\n\n\n\n

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.<\/strong><\/p>\n\n\n\n

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

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 \u00d7 5 + 2 \u00d7(5 + 3)]<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

The distances of potatoes from the bucket are 5, 8, 11, 14\u2026
Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept because first she has to first pick the potato and again return back to the same place in order to start picking the second potato.. Therefore, distances to be run are
10, 16, 22, 28, 34,\u2026\u2026\u2026.
a<\/em> = 10
d<\/em> = 16 \u2212 10 = 6
S<\/em>10<\/sub> =?
S10<\/sub> <\/em>= 10\/2 [2(20) + (n<\/em> – 1)(-1)]
= 5[20 + 54]
= 5 (74)
= 370
Therefore, the competitor will run a total distance of 370 m.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 5, class 10 Mathematics Chapter 5 solutions<\/p>\n\n\n\n

Exercise 5.4 (Optional)<\/h4>\n\n\n\n

1. Which term of the AP : 121, 117, 113, . . ., is its first negative term?
[Hint : Find n for an<\/sub> < 0]<\/strong><\/p>\n\n\n\n

Answer<\/strong>
We have the A.P. having a = 121 and d = 117 – 121 = – 4<\/p>\n\n\n\n

\u2234  an = a + (n – 1) d<\/p>\n\n\n\n

= 121 + (n – 1) \u00d7 (- 4)<\/p>\n\n\n\n

= 121 – 4n + 4<\/p>\n\n\n\n

= 125 – 4n<\/p>\n\n\n\n

For the first negative term, we have<\/p>\n\n\n\n

an < 0<\/p>\n\n\n\n

\u21d2 (125 – 4n) < 0<\/p>\n\n\n\n

\u21d2 125 < 4n<\/p>\n\n\n\n

\u21d2  125\/4 <n<\/p>\n\n\n\n

\u21d2 n > 31 1\u20444<\/p>\n\n\n\n

Thus, the first negative term is 32nd term.<\/p>\n\n\n\n

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

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3. A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 21\u20442 m apart, what is the length of the wood required for the rungs?<\/strong><\/p>\n\n\n\n

[Hint : Number of rungs = 250\/25 + 1]<\/p>\n\n\n\n

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Answer<\/strong><\/p>\n\n\n\n

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4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.<\/strong><\/p>\n\n\n\n

[Hint : Sx-1<\/sub><\/em> = S49<\/sub><\/em> \u2013 Sx<\/sub><\/em>]<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

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5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1\u20444 m and a tread of 1\u20442 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.<\/strong><\/p>\n\n\n\n

[Hint : Volume of concrete required to build the first step = 1\/4 \u00d7 1\/2 \u00d7 50m3<\/sup>]<\/p>\n\n\n\n

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Answer<\/strong><\/p>\n\n\n\n

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NCERT 10th Mathematics Chapter 5, class 10 Mathematics Chapter 5 solutions<\/p>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 5: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 5 – Arithmetic Progressions<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for Class 10th Mathematics: Chapter 5 – Arithmetic Progressions PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 10 Maths<\/strong>:<\/h4>\n\n\n\n