These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2

Question 1.

Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

Solution:

(i) 28

(ii) 2

(iii) 46

(iv) 10

(v) 3.5

Question 2.

Choose the correct choice in the following and justify:

(i) 30th term of the AP: 10, 7, 4, …, is

(A) 97

(B) 77

(C) -77

(D) -87

(ii) 11th term of the AP: -3, \(\frac { -1 }{ 2 }\) , 2, …, is

(A) 28

(B) 22

(C) -38

(D) -48

Solution:

(i) 10, 7, 4, …,

a = 10, d = 7 – 10 = – 3, n = 30

a_{n} = a + (n – 1)d

⇒ a_{30} = a + (30 – 1) d = a + 29 d = 10 + 29 (-3) = 10 – 87 = – 77

Therefore, 30th term of the sequence 10, 7, 4, is – 77 i.e., (C) is the correct choice.

(ii) We have given the sequence,

Therefore, 11th term of the sequence is 22 i.e., (B) is the correct choice.

Question 3.

In the following APs, find the missing terms in the boxes:

Solution:

Question 4.

Which term of the AP: 3, 8, 13, 18, …, is 78?

Solution:

Given: 3, 8, 13, 18, ………,

a = 3, d = 8 – 3 = 5

Let nth term is 78

a_{n} = 78

a + (n – 1) d = 78

⇒ 3 + (n – 1) 5 = 78

⇒ (n – 1) 5 = 78 – 3

⇒ (n – 1) 5 = 75

⇒ n – 1 = 15

⇒ n = 15 + 1

⇒ n = 16

Hence, a_{16} = 78

Question 5.

Find the number of terms in each of the following APs:

(i) 7, 13, 19, …, 205

(ii) 18, 15\(\frac { 1 }{ 2 }\), 13, …, -47

Solution:

(i) We have given the sequence 7,13,19,…. 205 Here, a = 7, d = 13-7 = 6 an = 205 Let n terms are in this AP.

∴ a_{n} = 205

a + (n – 1)d= 205 (∴ a_{n} = a + (n – 1) d)

or, 7 + (n – 1) x 6 = 205 (∴ a = 7 and d = 6)

or, (n – 1)6 = 205 – 7

or, (n – 1) = \(\frac { 198 }{ 6 }\)

∴ n = 33 + 1 = 34

Therefore, this sequence has 34 terms.

(ii) We have given the sequence

Question 6.

Check, whether -150 is a term of the AP: 11, 8, 5, 2, ….

Solution:

11, 8, 5, 2, …….

Here, a = 11, d = 8 – 11= -3, a_{n} = -150

a + (n – 1) d = an

⇒ 11 + (n – 1) (- 3) = -150

⇒ (n – 1) (- 3) = -150 – 11

⇒ -3 (n – 1) = -161

⇒ n – 1 = \(\frac { -161 }{ -3 }\)

⇒ n = \(\frac { 161 }{ 3 }\) + 1 = \(\frac { 164 }{ 3 }\) = 53\(\frac { 4 }{ 3 }\)

Which is not an integral number.

Hence, – 150 is not a term of the AP.

Question 7.

Find the 31^{st} term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:

a_{11} = 38 and a_{16} = 73

⇒ a_{11} = a + (11 – 1) d ⇒ a + 10d = 38 ….. (i)

⇒ a_{16} = a + (16 – 1 )d ⇒ a + 15d = 73 …(ii)

Subtracting eqn. (i) from (ii), we get

a + 15d – a – 10d = 73 – 38

⇒ 5d = 35

⇒ d = 1

From (i), a + 10 x 7 = 38

⇒ a = 38 – 70 = – 32

a_{31} = a + (31 – 1) d = a + 30d = – 32 + 30 x 7 = – 32 + 210 = 178

Therefore, the 31th term of this AP is 178.

Question 8.

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

Given:

a_{50} = 106

a_{50} = a + (50 – 1) d

⇒ a + 49d = 106 … (i)

and a_{3} = 12

⇒ a_{3} = a + (3 – 1 )d

⇒ a + 2d = 12 … (ii)

Subtracting eqn. (ii) from (i), we get

a + 49d – a – 2d = 106 – 12

⇒ 47d = 94

⇒ d = \(\frac { 94 }{ 47 }\) = 2

a + 2d = 12

⇒ a + 2 x 2 = 12

⇒ a + 4 = 12

⇒ a = 12 – 4 = 8

a_{29} = a + (29 – 1) d = a + 28d = 8 + 28 x 2 = 8 + 56 = 64

Therefore, the 29^{th} term of this AP is 64.

Question 9.

If the 3rd and the 9th term of an AP are 4 and -8 respectively, which term of this AP is zero?

Solution:

Given: a_{3} = 4 and a_{9} = – 8

⇒ a_{3} = a + (3 – 1 )d ⇒ a + 2d = 4 …(i)

a_{9} = a + (9 – 1) d ⇒ a + 8d = -8 ….(ii)

Subtracting eqn. (i) from (ii), we get

a + 8d – a – 2d = -8 – 4

⇒ 6d = -12.

⇒ d = -2

Now,

a + 2d = 4

⇒ a + 2(-2) = 4

⇒ a – 4 = 4

⇒ a = 4 + 4 = 8

Let a_{n} = 0

⇒ a + (n – 1) d = 0

⇒ 8 + (n – 1) (- 2) = 0

⇒ 8 = 2 (n – 1)

⇒ n – 1 = 4

⇒ n = 4 + 1 = 5

Therefore, 5th term of this AP is 0.

Question 10.

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution:

Given: a_{17} – a_{10} = 7

⇒ [a + (17 – 1 ) d] – [a + (10 – 1 ) d] = 7

⇒ (a + 16d) – (a + 9d) = 7

⇒ 7d = 7

⇒ d = 1

Therefore, the common difference of this AP is 1.

Question 11.

Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?

Solution:

3, 15, 27, 39, …..

Here, a = 3, d = 15 – 3 = 12

Let a_{n} = 132 + a_{54}

⇒ a_{n} – a_{54} = 132

⇒ [a + (n – 1) d] – [a + (54 – 1) d] = 132

⇒ a + nd – d – a – 53d = 132

⇒ 12n – 54d = 132

⇒ 12n – 54 x 12 = 132

⇒ (n – 54)12 = 132

⇒ n – 54 = 11

⇒ n = 11 + 54 = 65

Therefore, 65th term of this AP be 132 more than its 54th term.

Question 12.

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:

Let a and A be the first term of two APs and d be the common difference.

Given:

a_{100} – A_{100} = 100

⇒ a + 99d – A – 99d = 100

⇒ a – A = 100

⇒ a_{1000} – A_{1000} = a + 999d – A – 999d

⇒ a – A = 100

⇒ a_{1000} – A_{1000} = 100 [From equation (i)]

Therefore, difference of their 1000th term is 100.

Question 13.

How many three-digit numbers are divisible by 7?

Solution:

The three-digit numbers which are divisible by 7 are 105, 112, 119, ………., 994

Here, a = 105, d = 112 – 105 = 7 , a_{n} = 994

a + (n – 1) d = 994

⇒ 105 + (n – 1) 7 = 994

⇒ (n – 1) 7 = 994 – 105

⇒ 7 (n – 1) = 889

⇒ n – 1 = 127

⇒ n = 127 + 1 = 128

Therefore, the number of three digit numbers which are divisible by 7 is 128.

Question 14.

How many multiples of 4 lie between 10 and 250?

Solution:

The multiples of 4 between 10 and 250 be 12, 16, 20, 24,…., 248

Here, a = 12, d = 16 – 12 = 4, a_{n} = 248

a_{n} = a + (n – 1) d

⇒ 248 = 12 + (n – 1) 4

⇒ 248 – 12 = (n – 1) 4

⇒ 236 = (n – 1) 4

⇒ 59 = n – 1

⇒ n = 59 + 1 = 60

Therefore, number of terms between 10 and 250 which is multiple of 4 is 60.

Question 15.

For what value of n, the nth term of two APs: 63, 65, 61,… and 3, 10, 17,… are equal?

Solution:

First AP

63, 65, 67,…

Here, a = 63, d = 65 – 63 = 2

a_{n} = a + (n – 1) d = 63 + (n – 1) 2 = 63 + 2n – 2 = 61 + 2n

Second AP

3, 10, 17, …

Here, a = 3, d = 10 – 3 = 7

a_{n} = a + (n – 1) d = 3 + (n – 1)7 = 3 + 7n – 7 = 7n – 4

Now, a_{n} = a_{n}

⇒ 61 + 2n = 7n – 4

⇒ 61 + 4 = 7n – 2n

⇒ 65 = 5n

⇒ n = 13

Therefore, the 13th term of both APs are same.

Question 16.

Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.

Solution:

Given: a_{3} = 16

⇒ a + (3 – 1)d = 16

⇒ a + 2d = 16

and a_{7} – a_{5} = 12

⇒ [a + (7 – 1 )d] – [a + (5 – 1 )d] = 12

⇒ a + 6d – a – 4d = 12

⇒ 2d = 12

⇒ d = 6

Since a + 2d = 16

⇒ a + 2(6) = 16

⇒ a + 12 = 16

⇒ a = 16 – 12 = 4

a_{1} = a = 4

a_{2} = a_{1} + d = a + d = 4 + 6 = 10

a_{3} = a_{2} + d = 10 + 6 = 16

a_{4} = a_{3} + d = 16 + 6 = 22

Thus, the required AP is a_{1}, a_{2}, a_{3}, a_{4},…,

Therefore the required AP is 4, 10, 16, 22.

Question 17.

Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.

Solution:

Given: AP is 3, 8, 13,…….. ,253

On reversing the given A.P., we have

253, 248, 243 ,………, 13, 8, 3.

Here, a = 253, d = 248 – 253 = -5

a_{20} = a + (20 – 1)d = a + 19d = 253 + 19 (-5) = 253 – 95 = 158

Question 18.

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

We have,

or a_{4} + a_{8} – 24

a + 3d + a + 7d = 24

or 2a + 10d = 24 … (i)

and a_{6} + a_{10}

or a + 5d + a + 9d = 44

or 2a + 14d = 44 … (ii)

Subtracting equation (ii) from equation (i), we

2a + 10d – (2a + 14d) = 24 – 44

or 2a + 10d – 2a – 14d = – 20

or 10d – 14d = – 20

or – 4d = – 20

∴ d = \(\frac { -20 }{ -4 }\)

Putting the value of d in equation (i), we get

2a + 10d = 24

or 2a + 10 x 5 = 24

or 2a + 50 = 24

or 2a = 24 – 50

or 2 a = – 26

∴ a = \(\frac { -26 }{ 2 }\) = – 13

Therefore, the sequence can be,

– 13, – 13 + 5, – 13 + 2 x 5,… or -13, -8, -3,…

∴ The first three terms of this AP are -13, -8, and-3

Question 19.

Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000 ?

Solution:

We have,

Starting salary of Subba Rao is ₹ 5000 and annual increment = ₹ 500

Therefore, the sequence of Subba Rao’s salary is 5000, 5200, 5400,…

We have, a + (n – 1) d = 7000

⇒ 5000 + (n – 1) 200 = 7000

⇒ (n – 1) 200 = 7000 – 5000

⇒ (n – 1) 200 = 2000

⇒ (n- 1) = 10

⇒ n = 11

⇒ 1995 + 11 = 2006

Therefore, after 11 years from 1995, salary of Subba Roa each ₹ 7000.

∴ 19995 + 11 = 2006

So, in year 2006, salary of Subba Rao will reach will ₹ 7000.

Question 20.

Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving become ₹ 20.75, find n.

Solution:

Ramkali savings in the first week of the year be ₹ 5.

and weekly saving increased by ₹ 1.75

Therefore, the sequence of weekly savings be 5, 6.75,8.50,…

Here, a = 5 and d = 6.75 – 5 = 1.75

Let n^{th} term of this sequence be 20.75

a_{n} = 20.75

a + (n – 1) d – 20.75

⇒ 5 + (n – 1) 1.75 = 20.75

⇒ (n – 1) x 1.75 = 20.75 – 5

⇒ (n – 1) 1.75 = 15.75

⇒ n – 1 = 9

⇒ n = 9 + 1

⇒ n = 10

Hence, in 10th week Ramkali’s saving will be ₹ 20.75.