Download NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2 – Arithmetic Progressions. This Exercise contains 20 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

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

**NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2 – Arithmetic Progressions**

**A. Fill in the blanks in the following table, given that a is the first term, d the common difference and a _{n} the n^{th} term of the AP:**

**Solution :** If a be the first term and d be the common difference of an AP then the n^{th} term of that AP is given by

a_{n} = a+(n-1)d

**(i) a = 7, d = 3, n = 8**

**Solution:** We know that,

a_{n} = a + (n-1) d

= 7 + (8-1)×3

= 7 + 21

= 28

**(ii) a = -18, n = 10, a _{n} = 0**

**Solution:** We have,

a_{n} = a + ( n – 1 ) d

Substituting the given values in the above formula we get,

0 = -18 + (10-1) d

0 = -18 + 9d

18 =9d

d = 2

**(iii) d = -3, n =18, a _{n}= -5**

**Solution:** We know that,

a_{n} = a + (n-1) d

Substituting the given values in the above formula we get

– 5 = a + (18 – 1)×(- 3)

– 5 = a + 17 ×(- 3)

– 5 = a-51

a = 51-5

a = 46

**(iv) a = – 18.9, d = 2.5, a _{n} = 3.6**

**Solution:** We know the relation is

a_{n}= a + (n – 1)d

Substituting the given values in the above equation,

3.6 = -18.9+(n – 1) ×2.5

(n – 1)×2.5 = 3.6+18.9

(n – 1) ×2.5 = 22.5

n – 1= (22.5/2.5)

n – 1 = 9

n = 9 + 1

n = 10

**(v) a = 3.5, d = 0, n =105**

**Solution:** We know that,

a_{n} = a + (n – 1)d

Substituting the values in the above equation

a_{n} = 3.5 +( 105-1)×0

= 3.5 + 104×0

= 3.5 + 0

= 3.5

**Answer**

**B. Choose the correct choice in the following and justify :**

**(i) 30 ^{th} term of the AP: 10, 7, 4, . . . , is**

**(A) 97 (B) 77 (C) –77 (D) – 87**

** ****Solution:** The AP is given by 10, 7, 4, …

First term, a = 10,

Common difference, d = 7 – 10 = – 3

We know that,

a_{n} = a + (n – 1) d

a_{30} = a + (30 – 1)×d

a_{30}= 10 + (30 – 1)(- 3)

a_{30}= 10 + 29× (-3)

a_{30}= 10 – 87

a_{30}= -77

Hence, the correct answer is option (C)

**(ii) 11 ^{th} term of the AP: – 3,**

**(-1/2)**

**,**

**2, . . ., is**

**(A) 28 (B) 22 (C) –38 (D) –****48**** **

**Solution :** AP is given by -3, (-1/2), 2, …

First term, a = -3

Common difference, d = (-1/2) – ( – 3 )

= (-1/2) + 3 = (5/2)

We know that,

a_{n} = a + (n – 1) d

a_{11}= -3+(11-1)(5/2)

a_{11}= – 3+10×(5/2)

a_{11}= – 3+25

a_{11}= 22

Hence, the correct answer is option (B)

**3. In the following APs find the missing terms in the boxes:**

**(i) 2, ___, 26**

**Solution: **In the given A.P.,

a = 2

a_{3} = 26

We know that,

a_{n} = a + (n – 1) d

a_{3} = 2 + (3 – 1) d

26 = 2 + 2d

24 = 2d

d = 12

a_{2} = a + d = 2 + 12 = 14

Hence the missing term is 14.

**(ii) ___, 13, ___, 3**

**Solution:** In the given AP, a_{2} =13 and a_{4} = 3

We know that,

a_{n} = a + (n – 1) d

a+d = 13….(1) and

a+3d = 3….(2)

On Subtracting equation (1) from equation (2), we get

2d = -10

d = -5

a = a_{2} – d = 13 – (-5) = 13 + 5 = 18

a_{3} = a_{2} + d = 13 + (-5) = 13 – 5 = 8

Hence the missing term are 18 and 8.

**(iii) 5, ___, ___, 9 1⁄2**

**Solution:** In the given AP, a = 5 and a_{4} = 9 1⁄2

We know that,

a_{n} = a + (n – 1) d

a_{4}= a+(4-1)d

9 1⁄2 = 5+3d

3d = (19/2) -5

3d = (19/2)

d = 3/2

a_{2} = a + d = 5 + (3/2) = (13/2) = 6 1⁄2

a_{3} = a+2d = 5+ 2 3⁄2 = 5+3 = 8

Hence the missing terms are 6 and 8 respectively.

**(iv) -4 , __, __, __, 6**

**Solution:** In the given AP, a = -4 and a_{4} = 6

We know that,

a_{n} = a + (n – 1) d

a_{6}= a+(6-1)d

6 = -4+5d

5d = 10

d = 2

Therefore,

a_{2} = a + d = -4+2 = -2

a_{3} = a+2d = -4+2×2 = -4+4 = 0

a_{4 }= a+3d = -4+3×2 = -4+6 = 2

a_{5 }= a+4d = -4+4×2 = -4+8 = 4

Therefore the missing terms are -2, 0, 2, 4

**(v) __, 38, __, __, __, -22**

**Solution: **In the given AP, a_{2} = 38 and a_{6} = -22

We know that,

a_{n} = a + (n – 1) d

a_{2}= a+(2-1)d

38 = a+d

a+d = 38…..(1)

a_{6 }= a+(6-1)d

-22 = a+5d

a+5d = -22…..(2)

On Subtracting equation (1) from equation (2) we get

4d = -60

d = -15

Substituting the value of d in the equation (1) we get

a-15 = 38

a = 53

Therefore

a_{3} = a+2d = 53+2×(-15) = 53-30 = 23

a_{4 }= a+3d = 53+3×(-15) = 53-45 = 8

a_{5 }= a+4d = 53+4×(-15) = 53-60 = -7

Therefore the missing terms are 53, 23, 8, -7 respectively.

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

**Solution:** Let 78 is the n^{th} term of the given AP.

First term, a = 3

Common ratio, d = 8-3 = 5

We know that,

a_{n} = a + (n-1) d

78 = 3+(n-1)5

75 = (n-1)5

15 = n-1

n = 16

Hence, 16^{th} term is of the given AP is 78.

**5. Find the number of terms in each of the given APs.**

**(i) 7, 13, 19, …,205**

**Solution:** Let the number of terms of the given AP is n.

First term, a =7

Common difference, d = 13-7 = 6.

a_{n} = 205

a_{n} = a + (n-1) d

205 = 7+(n-1)6

198 = (n-1)6

n-1 = 33

n = 34

Hence, the number of terms of the AP is 34.

**(ii) 18, 15 1⁄2 ****, 13, …,.-47**

**Solution:** Let the number of terms of the given AP is n.

First term, a =18

Common difference, d = 15 1⁄2 – 18 = (31/2) – 18 = (-5/2)

As we know that,

a_{n} = a + (n-1) d

-47 = 18+(n-1)(-5/2)

-65 = (n-1)(-5/2)

(65×2)/5= n-1

26 = n -1

n = 27

Hence, the number of terms of the AP is 27.

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

**Solution:** In the given AP,

First term, a = 11

Common difference d = 8-11 = -3

Let n^{th} term of the given AP is -150.

As we know that,

a_{n} = a + (n-1) d

-150 = 11+(n-1)(-3)

-161 = (n-1)(-3)

n-1 = (161/3)

n = 1+(161/3), which is not possible because 1+(161/3) is not an integer value.

Hence, -150 is not a term of this AP.

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

**Solution:** Let a be the first term and d be the common difference of the AP.

As we know that,

a_{n} = a + (n-1) d

a_{11} = a + (11-1) d

38 = a + 10 d

a+10d = 38……(1) and

Similarly, a_{16} = a + (16-1) d

73 = a+15d

a+15d = 73……(2)

Subtracting the equation (1) from the equation (2) we get

5d = 35

d = 7

Substituting the value of d in the equation (1) we get

a+10×7 = 38

a+70 = 38

a = -32

Then, 31th term of the AP is

a_{31} = -32 + (31-1) 7

a_{31} = -32+210

a_{31 }= 178.

Hence, 31^{st} term of the AP is 178.

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

**Solution:** Let a be the first term and d be the common difference of the AP.

As we know that,

a_{n} = a + (n-1) d

a_{3} = a + (3-1) d

12 = a+2d

a+2d = 12……(1) and

a_{50} = a + (50-1) d

106 = a+49d

a+49d = 106……(2)

Subtracting the equation (1) from (2) we get

47d = 94

d = 2

Substituting the value of d in the equation (1) we get

a+2×2 = 12

a+4 = 12

a = 8

Then, 29^{th} term of the AP is,

a_{29} = 8+(29-1)2

a_{29 }= 8+28×2

a_{29 }= 64.

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

**9. If the 3 ^{rd} and 9^{th} terms of an AP are 4 and -8 respectively, which term of this AP is zero?**

**Solution:** Let a be the first term and d be the common difference of the AP and n^{th} term of this AP be zero.

Given:

a_{3 }= 4

a_{9 }= -8

As we know that,

a_{n} = a + (n-1) d

a_{3} = a + (3-1) d

4 = a+2d

a+2d = 4……(1)

a_{9} = a + (9-1) d

-8 = a + 8d

a+8d = -8……(2)

Subtracting the equation (1) from equation (2) we get

6d = -12

d = -2

Substituting the value of d in the equation (1) we get

a+2(-2) = 4

a-4 = 4

a = 8

If n^{th} term of this AP is zero, then a_{n} = 0

As we know that,

a_{n} = a + (n-1) d

0 = 8+(n-1)(-2)

-(n-1)2 = -8

n-1 = 4

n = 5

Therefore the 5^{th} term of this AP is zero.

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

**Solution:** Let a be the first term and d be the common difference of the AP.

As we know that,

a_{n} = a + (n-1) d

a_{17} = a + (17-1) d

a_{17} = a + 16d

Similarly, a_{10} = a + 9d

Then using the given condition we have,

a+16d = a+9d+7

7d = 7

d = 1

Therefore, the common difference of the AP is 1.

**11. Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54 ^{th} term**

**Solution:** In the given AP,

First term, a = 3

Common difference, d = 15-3 = 12.

As we know that,

a_{n} = a + (n-1) d

a_{54} = 3 + (54-1) 12

a_{54} = 3 + 53 x 12

a_{54} = 3 + 636

a_{54} = 639

Let n^{th} term of the given AP be 132 more than 54^{th} term.

3+(n-1)12 = a_{54} + 132

3+(n-1)12 = 639 + 132

3+(n-1)12 = 771

(n-1)12 = 768

n-1 = 64

n = 65.

Hence, 65^{th} term of this AP is 132 more than 54^{th} term.

**12. Two APs has the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000 ^{th} terms?**

**Solution:** Let a and b be the first terms of the given two APs respectively and the common difference of both the APs is d.

As we know that,

a_{n} = a + (n-1) d

For 1^{st }AP

a_{100} = a + (100-1) d

a_{100} = a + 99d

a_{1000} = a + (1000-1) d

a_{1000} = a + 999d

For 2^{nd} AP

a_{100} = b + (100-1) d

a_{100} = b + 99d

a_{1000} = b + (1000-1) d

a_{1000} = b + 999d

Then using the given condition, we have

(a+99d)-(b+99d) = 100

a-b = 100 —— (Eq 1)

Difference between their 1000^{th} terms,

= (a+999d)-(b+999d)

= a-b (From eq 1)

= 100

Hence, the difference between 1000^{th} terms of this AP is 100.

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

**Solution:** We know that the first and last three-digit numbers divisible by 7 are 105 and 994.

Therefore, we get an AP whose first term is 105, common difference is 7 and last term is 994. Let 994 is the n^{th} term.

a = 105

d = 7

a_{n} = 994

n = ?

As we know that,

a_{n} = a + (n-1) d

994 = 105 + (n-1)7

889 = (n-1)7

127 = n-1

n = 128

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

**14. How many multiplies of 4 lies between 10 and 250?**

**Solution:** We know that the numbers divisible by 4 after 10 and before 250 are

12, 16, 20, …, 248

First term, a = 12

Common difference, d = 16-12 = 4

Let the n^{th} term of this AP is 248.

a_{n }= 248

As we know that,

a_{n} = a + (n-1) d

248 = 12+(n-1)4

236 = (n-1)4

n-1 = 59

n = 60

There are 60 multiples of 4 between 10 and 250.

**15. For what value of n, are the n ^{th} term of two APs: 63, 65, 67,…and 3, 10, 17,…equal?**

**Solution:** For the AP: 63, 65, 67,…

First term, a = 63

Common difference, d = 65-63 = 2.

a_{n} = 63+(n-1)2

And for the AP 3, 10, 17, …

First term, a = 3

Common difference, d = 10-3 = 7.

a_{n} = 3+(n-1)7

Then by given condition we have

63+(n-1)2 = 3+(n-1)7

5(n-1) = 60

n-1 = 12

n = 13.

Therefore, 13^{th} term of both the A.P.s are equal.

**16. Determine the AP whose third term is 16 and the 7 ^{th} term exceed the 5^{th} term by 12.**

**Solution: **Let a be the first term and d be the common difference of the AP.

a_{3 }= a + (3-1)d

16 = a + 2d ——–(1)

a_{7 }= a + (7-1)d

a_{7 }= a + 6d

a_{5 }= a + (5-1)d

a_{5 }= a + 4d

a_{7 }– a_{5 }= 12

a + 6d – (a + 4d) = 12

a + 6d – a – 4d = 12

2d = 12

d = 6

Substituting the value of d in the equation (1) we get

16 = a+2×6

16 = a+12

a = 4

Therefore, A.P. will be

a, a+d, a+2d,…

4, 4+6, 4+2×6,…

4, 10, 16,…

**17. Find the 20 ^{th} term from the last term of the AP: 3, 8, 13, …, 253.**

**Solution:** A.P. = 3, 8, 13,……, 253.

To find the 20^{th} term from the last of the A.P., we will reverse the order of A.P.

253, …….13, 8, 3

First term, a = 253

Common difference, d = 3-8 = -5

n = 20

a_{n }= ?

As we know that,

a_{n} = a + (n-1) d

a_{n} = 253+(20-1)-5

a_{n} = 253 +19 x -5

a_{n }= 253 – 95

a_{n }= 158

Therefore the 20^{th} term from the last term is 158

**18. The sum of the 4 ^{th} and 8^{th} term of the AP is 24 and the sum of the 6^{th} and 10^{th} term is 44. Find the first three terms of the AP.**

**Solution:** As we know that,

a_{n} = a + (n-1) d

a_{4} = a + (4-1) d

a_{4} = a + 3d

Similarly,

a_{8} = a + 7d

a_{6} = a + 5d

a_{10} = a + 9d

From the given condition,

(a+3d)+(a+7d) = 24

2a+10d = 24……(1) and

(a+5d)+(a+9d) = 44

2a+14d = 44……(2)

Subtracting the equation (1) form equation (2) we get

4d = 20

d = 5

Substituting the value of d in equation (2), we get

2a+10×5 = 24

2a+50 = 24

2a = -26

a = -13

a_{2} = a + d = -13 + 5 = -8

a_{3} = a + 2d = -13 + 2(5) = -3

Therefore, first three terms of the AP are given by -13, -8, -3.

**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:** Since the increment of each year is same. Therefore, it forms an AP,

5000, 5200, 5400, ……….7000

Here, a = 5000

d = 200

a_{n} = 7000

n = ?

We know that,

a_{n} = a + (n-1)d

7000 = 5000+(n-1)200

2000 = (n-1)200

n-1 = 10

n = 11

Therefore, in 11 year his salary will be ₹7000.

**20. Ramkali save ****₹****5 in the first week of a year and then increased her weekly savings by ****₹****75. If in the n ^{th} week, her weekly savings become **

**₹**

**20.75, find n.**

**Solution:** Since the increment is same, therefore it forms an AP

a = 5

d = 1.75

a_{n }= 20.75

We know that,

a_{n} = a + (n-1)d

20.75 = 5+(n-1)×1.75

15.75 = (n-1)×1.75

9 = n-1

n =10

Therefore, the required value of n is 10.

**NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.2 – Arithmetic Progressions, has been designed by the NCERT to test the knowledge of the student on the topic – nth Term of an AP**

**Download NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2 – Arithmetic Progressions**