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Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions – Exercise 5.2

  • Last Updated : 03 Mar, 2021

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 A.P

 adnan
I738
II-18100
III-318-5
IV-18.92.53.6
V3.50105

Solution: 

(i) a = 7, d = 3, n = 8 

Using A.P formula, 

an = a + (n – 1)d

So, an = 7 + (8 – 1) * 3



an = 7 + 21

an = 28

(ii) a = -18, d = ?, n = 10, an = 0

Using A.P formula, 

an = a + (n – 1)d

0 = -18 + (10 – 1) * d

18 = 9 * d

d = 2

(iii) a = ?, d = -3, n = 18, an = -5

Using A.P formula, 

an = a + (n – 1)d

-5 = a + (18 – 1) * (-3)

-5 = a – 51

a = -5 + 51

a = 46

(iv) a = -18.9, d = 2.5, n = ?, an = 3.6 

Using A.P formula, 

an = a + (n – 1)d



3.6 = -18.9 + (n – 1) * 2.5

3.6 + 18.9 = (n – 1) * 2.5

n – 1 = 22.5/2.5

n – 1 = 9

n = 10

(v) a = 3.5, d = 0, n = 105, an = ?

Using A.P formula, 

an = a + (n – 1)d

an = 3.5 + (105 – 1) * 0

an = 3.5

Question 2. Choose the correct choice in the following and Justify

(i) 30th term of the A.P: 10, 7, 4, ….., is 

(A) 97          (B) 77         (C) -77          (D) -87

Solution: 

Given: A.P. is 10, 7, 4, ……

Now we find the common difference:

Common difference(d) = second term – first term 

So, d = 7 -10 = -3

The first term (a) = 10

Total number of terms (n) = 30

Now, we find the 30th term

a30 (30th term) = a + (n – 1)d

a30 = 10 + (30 – 1) * (-3)



a30 = 10 – 87

a30 = -77

Hence, the correct option is C.

(ii) 11th term of the AP -3, -1/2, 2, ……. is 

(A) 28          (B) 22          (C) -38          (D) -48\tfrac{1}{2}

Solution: 

Given: A.P. is -3, -1/2, 2, …….

So, first term (a) = -3

Now, we find the common difference;

Common Difference (d) = second term – first term 

d = -1/2 – (-3)

d = -1/2 + 3

d = 5/2

11th term can be calculated by following formula

an = a +(n – 1)d

a11 = -3 + (11 – 1) * (5/2)

= -3 + (10) * (5/2)

= -3 + 25

a11 = 22

Hence, option B is the correct choice.

Question 3. In the following APs, find the missing term in the boxes 

(i) 2, \square, 26

Solution : 

Given: first term (a) = 2



third term (a3) = 26

a3 can be calculated using the formula an = a + (n – 1)d

a3 = 2 + (3 – 1) * d   

26 = 2 + 2d

24 = 2d

d = 12

So a2 can be calculated using the formula an = a + (n – 1)d

a2 = 2 + (2 – 1) * 12

a2 = 2 + 12

a2 = 14

(ii)\square, 13, \square, 3

Solution: 

Given:

a2 = 13

a + (2 – 1)d = 13

a + d = 13         -(1)

a4 = 3

a + (4 – 1)d = 3

a + 3d = 3         -(2)

After solving equation (1) and (2), you will get 

d = -5

and a = 18

Now we find a3

a3 = 18 + (3 – 1) * (-5)

a3 = 18 -10

a3 = 8

Hence, the missing terms in the square boxes are 18 and 8.

(iii) 5, \square\square9\tfrac{1}{2}

Solution: 

Given:

a  = 5

a4 = 19/2



a + (4 – 1)d = 19/2

5 + 3d = 19/2

3d = (19/2) – 5

3d = 9/2

d = 9/6 = 3/2

So, a2 = 5 + (2 – 1) * (3/2)

a2 = 5 + 3/2

a2 = 13/2

and a3 = 5 + (3 – 1) * (3/2)

a3 = 5 + 3

a3 = 8

Hence, the missing terms in the square boxes are 13/2 and 8.

(iv) -4, \square\square\square\square, 6

Solution: 

Given: a = -4

a6 = 6

a + (6 – 1)d = 6

-4 + 5d = 6

5d = 10

d = 2

So, a2 = a + d

a2 = -4 + 2 = -2

a3 = a + 2d

a3 = -4 + 4 = 0

a4 = a + 3d

a4 = -4 + 6 = 2

a5 = a + 4d

a5 = -4 + 8 = 4

Hence, the missing terms in the square boxes are -2, 0, 2, and 4.

(v) \square, 38, \square\square\square, -22

Solution: 

Given:



a2 = 38

a + d = 38         -(1)

a6 = -22

a + 5d = -22         -(2)

After solving (1) and (2) equation, you will get

d = -15

and a= 53

So,

a3 = a + 2d 

a3 = 53 – 30 = 23

a4 = a+ 3d

a4 = 53 – 45 = 8

a5 = a + 4d 

a5 = 53 – 60 = -7

Hence, the missing terms in the square boxes are 53, 23, 8 and -7.

Question 4. Which term of the A.P. 3, 8,13, 18, ……. is 78 ?

Solution: 

Given:

a = 3

d = second term – first term

d = 8 – 3 = 5

and an = 78

n = ?

a + (n – 1)d = 78

3 + (n – 1) * 5 = 78

(n – 1) * 5 = 78 – 3 

n – 1 = 75 /5

n = 15 +1

n = 16

So, 78 is the 16th term of the given A.P.

Question 5. Find the number of terms in each of the following A.P.

(i) 7,13, 19, ……, 205

Solution: 

Find: n = ?

Given: a = 7

d = 13 – 7 = 6

an = 205

a + (n – 1)d = 205

7 + (n – 1) * 6 = 205

(n – 1) * 6 = 205 – 7

n – 1 = 198/6

n = 33 + 1

n = 34

So, there are 34 terms in the given A.P.

(ii) 18, 15\tfrac{1}{2}, 13, ……., -47

Solution: 

Find: n = ?

Given: a = 18

d = 31/2 – 18 = -5/2

an = -47 

18 + (n – 1) * (-5/2) = -47

(n – 1) * (-5/2) = -47 – 18

n – 1 = -65 * (-2/5)

n – 1 = 130/5

n = 26 + 1

n = 27

Hence, there are 27 terms in the given A.P.

Question 6. Check whether -150 is a term of the A.P. 11, 8, 5, 2,……..

Solution: 

Given:

a = 11

d = 8 – 11 = -3

Suppose -150 is the nth term of the given A.P. then it can be written as

a + (n – 1)d = -150

11 + (n – 1) * (-3) =- 150



(n – 1) * (-3) = -150 – 11

n – 1 = -161/(-3)

n = 161/3 + 1

n = 164/3

As you can see that n is not in integer. Hence, -150 is not a term in the given A.P.

Question 7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Solution: 

Given:

a11 = 38

a + 10 * d = 38         -(1)

a16 = 73

a + 15 * d = 73         -(2)

Subtracting equation (1) from equation (2), you will get

15 * d – 10 * d = 73 – 38

d = 35/5

d = 7

After putting value of d in equation (1) you will the value of a

a = 38 – 70

a = -32

So, a31 = a + (31 – 1) * d

a31 = -32 + 30 * 7

a31 = 178

Hence, 31st term of the given A.P. is 178.

Question 8. In A.P. consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.

Solution

According to the question:

n = 50

a3 = 12

a + 2 * d = 12         -(1)

a50 = 106

a + 49 * d  = 106         -(2)

After solving (1) and (2) you will get

47 * d = 106 – 12

d = 94/47 = 2

and a = 12 -4 = 8

So, a29 = a + 28 * d

a29 = 8 + 28 * 2

a29 = 64

So the 29th term of the given A.P. will be 64.

Question 9. If the 3rd and the 9th terms of an A.P. are 4 and -8 respectively. Which term of this A.P. is zero?

Solution: 

Given: a3 = 4

a + 2 * d = 4         -(1)

and a9 = -8

a + 8 * d = -8          -(2)

After solving (1) and (2), you will get 

d = -2 

and a = 8

Let the nth term of the A.P. will be zero.

So an = 0

a + (n – 1)d = 0

8 + (n – 1) * (-2) = 0

n – 1 = -8/(-2)

n = 4 + 1

n = 5

Hence, the 5th term of the given A.P. will be zero.

Question 10. If the 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Solution: 

According to the question

a17 = a10 + 7 

a + (17 – 1)d = a + (10 – 1)d + 7

16 * d = 9 * d + 7

7 * d = 7 

d = 1

Hence, the common difference of the given A.P. will be 1.

Question 11. Which term of the A.P. 3, 15, 27, 39, ….. will be 132 more than its 54th term?

Solution: 

According to the question

d = 15 – 3 = 12

Let the nth term of the A.P. will be 132 more than its 54th term

an = a54 + 132

a + (n – 1)d = a + (54 – 1)d + 132

(n – 1)(12) = 53 * 12 + 132

(n – 1) * 12 = 636 + 132

n – 1 = 771 / 12



n = 64 + 1

n = 65

Hence, 65th term will be 132 more than its 54th term.

Question 12. Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th term?

Solution: 

Let d be common difference of both APs and a1 be the first term of 

one A.P. and a2 be the first term of other A.P.

So, a100 for first A.P. will be 

a100 = a1 + 99d 

and a100 for second A.P. will be 

a100 = a2 + 99d

According to the question, the difference between their 100th term is 100

So, (a1 + 99d) – (a2 + 99d) = 100

a1 – a2 = 100         -(1)

Now a100 for first A.P. will be

a1000 = a1 + 999d

and a1000 for second A.P. will be

a1000 = a2 + 999d

Difference between their 1000th term will be

= (a1 + 999d) – (a2 + 999d)

= a1 – a2

= 100             -(from (1) a1 -a2 = 100)

Hence, the difference between their 1000th term will be 100.

Question 13. How many three-digit numbers are divisible by 7?

Solution: 

First three-digit number that is divisible by 7 = 105

So, the first term of the AP (a) = 105

and common difference (d) = 7

The last three-digit number that is divisible by 7 = 994

So AP will look like 105, 112, 119,………, 994

Let there are n three-digit numbers between 105 and 994

You have an = 994

a + (n – 1)d = 994

105 + (n – 1) * 7 = 994

(n – 1) * 7 = 994 – 105

n – 1 = 889/7

n = 127 + 1

n = 128

Hence, there 128 three-digit numbers that are divisible by 7.

Question 14. How many multiples of 4 lie between 10 and 250?

Solution: 

Given:

12 is the minimum number that is divisible by 4 between 10 and 250.

So, a = 12

d = 4 

248 is the highest number that is divisible by 4 between 10 and 250.

So an = 248

a + (n – 1)d = 248

12 + (n – 1) * 4 = 248

(n – 1) * 4 = 236

n – 1 = 59

n = 60

Hence, there are 60 multiples of 4 between 10 and 250.

Question 15. For what value of n, are the nth terms of two APs 63, 65, 67,…. and 3, 10, 17,….. equal?

Solution: 

For A.P. 63, 65, 67,…..

a = 63 and d = 65 – 63 = 2

nth term for this AP will be

an = 63 + (n – 1) * 2

For A.P. 3, 10, 17, ….

a = 3 and d = 10 – 3 = 7

nth term for this AP will be

an = 3 + (n – 1) * 7

According to the question, nth terms of both APs are equal

So, 63 + (n – 1) * 2 = 3 + (n – 1) * 7

60 = (n – 1) * 5

n – 1 = 12

n =13

Hence, 13th of both the APs are equal.

Question 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution: 

a3 = 16

a + 2d = 16          -(1)

and a7 = a5 + 12

a + 6d = a + 4d + 12



2d = 12

d = 6

On putting value of d in equation (1), you will get

a = 16 – 12

a = 4

So, AP will look like, 4, 10, 16, 22, 28,…..

Question 17. Find the 20th term from the last term of the AP 3, 8, 13,………, 253

Solution: 

d = 8 – 3 = 5

In reverse order the AP will be

253, 248, ………. 13, 8, 3

Now for this AP

a = 253 and d = 248 – 253 = -5

So 20th term will be

a20 = a + 19d

a20 = 253 + 19 * (-5)

a20 = 158

Hence, the 20th term from the last term for the given AP will be 158.

Question 18. The sum of 4th and 8th terms of an AP is 24 and the sum of the 6th term and 10th terms is 44. Find the first three terms of the AP.

Solution:

Given: a4 + a8 = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12         -(1)

and a6 + a10 = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22         -(2)

From (1) and (2), you will get

d = 5

and a = -13

So a2 = a + d

a2 = -13 + 5 = -8

a3 = a + 2d

a3 = -13 + 10 = -3

Hence, the first three terms of the AP are -13, -8, -3

Question 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.

 Solution: 

As, salary of Subba Rao is increasing by a fixed amount in every year hence 

this will form an AP with first term (a) = 5000 and common difference (d) d = 200

an = 7000

a +(n – 1)d = 7000

5000 + (n – 1) * 200 = 7000

n – 1 = 2000/200

n = 10 + 1

n = 11

Hence, the salary will be 7000 in 11th year.

Question 20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by 1.75. If in the nth week her weekly savings become Rs 20.75, find n.

Solution: 

As saving is increased by a fixed amount, this will form an AP in which

First term (a) = 5

Common difference (d) = 1.75

an = 20.75

a + (n – 1)d = 20.75

5 + (n – 1) * (1.75) = 20.75

(n – 1) * (1.75) = 20.75 – 5

n – 1 = 15.75/1.75

n = 9 + 1

n = 10

Hence, n = 10




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