**Download 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=

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,**

**,**

**2, . . ., is**

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

**Solution :** AP is given by -3, , 2, …

First term, a = -3

Common difference, d = – ( – 3 )

= + 3 =

We know that,

a_{n} = a + (n – 1) d

a_{11}= -3+(11-1)

a_{11}= – 3+10×

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, , , **

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

We know that,

a_{n} = a + (n – 1) d

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

= 5+3d

3d = -5

3d =

d =

a_{2} = a + d = 5 + = =

a_{3} = a+2d = 5+ = 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, ****, 13, …,.-47**

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

First term, a =18

Common difference, d = – 18 = – 18 =

As we know that,

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

-47 = 18+(n-1)

-65 = (n-1)

= 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 =

n = 1+, which is not possible because 1+ 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.

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

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