# RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progression Exercise 9.4

**1. ****Find:**

**(i) 10th term of the A.P. 1, 4, 7, 10, .......(ii) 18th term of A.P. √2, 3√2, 5√2, .......(iii) nth term of the A.P. 13, 8, 3, -2, ......(iv) 10th term of the A.P. -40, -15, 10, 35, .....(v) 8th term of the A.P. 11, 104, 91, 78, .......(vi) 11th term of the A.P. 10.0, 10.5, 11.0, 11.5, .....(vii) 9th term of the A.P. 3/4, 5/4, 7/4, 9/4 , .......**

**Solution**

(i) Given A.P. is

1, 4, 7, 10, ......

First term (a) = 1

Common difference (d) = second term - first term

= 4 - 1 = 3

nth term in an A.P = a (n -1)d

10th term in an A.P = 1 + (10 - 1)3

= 1 + 9×3

= 1 + 27 = 28

(ii) Given, A.P is √2, 3√2, 5√2, .......

First term (a) = √2

common difference(d) = second term - first term

3√2 - √2 = 2√2

n th term in an A.P = a + (n -1)d

18 th term of A.P = √2 + (18 - 1)2√2

= √2 + 17 × 2√2

= √2(1 + 34)

= 35√2

(iii) Given, A.P is 13, 8, 3, -2, .......

First term (a) = 13

Common difference (d) = Second term - first term

8 - 13 = - 5

nth term of an A.P = a + (n -1)d

= 13 + (n -1) - 5

= 13 - 5n + 5

= 18 - 5n

(iv) Given A.P is

-40, -15, 10, 35, .......

First term (a) = -40

Common difference (d) = Second term - first term

= -15 - (-40)

= 40 - 15 = 25

nth term of an A.P = a + (n -1)d

10th term of A.P = -40 + (10 - 1)25

= -40 + 9×25

= -40 + 225 = 185

(v) Given sequence is

117, 104, 91, 78, .........

First learn can = 117

Common difference (d) = Second term - first term

= 104 - 117 = - 13

nth term = a + (n -1)d

8th term = a + (8 - 1)d

= 117 + 7(-13)

= 117 - 91 = 26

(vi) Given A.P is 10.0, 10.5, 11.0, 11.5, .....

First term (a) = 10.0

Common difference (d) = Second term - First term

= 10.5 - 10.0 = 0.5

nth term = a + (n - 1)d

11th term = 10.0 + (11 - 1) 0.5

= 10.0 + 10 ×0.5

= 10.0 + 5= 15.0

(vii) Given A.P. is 3/4, 5/4, 7/4, 9/4 , ........

First term (a) = 3/4

Common difference (d) = Second term - First Term

= 5/4 - 3/4 = 2/4

nth term = a + (n - 1)d

9th term = a + (9 -1)d

= 3/4 + 8× 2/4

= 3/4 + 16/4

= 19/4

**2. (i) Which term of the AP 3, 8, 13, ..... is 248?(ii) Which term of the AP 84, 80, 76, ...... is 0 ?(iii) Which term of the AP 4, 9, 14, ...... is 254 ? (iv) Which term of the AP 21, 42, 63, 84, ...... is 420? (v) Which term of the AP 121, 117, 113, ...... is its first negative term ? **

**Solution**

(i) Given A.P is 3, 8, 13, .....

First term (a) = 3

common difference (d) = Second term - first term

= 8 - 3 = 5

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

Given nth term a_{n} = 248

248 = 3 + (n - 1)5

⇒ 248 = -2 + 5n

⇒ 5n = 250

⇒ n = 250/5 = 50

50th term is 248

(ii) Given A.P is 84, 80, 76, .......

First term (a) = 84

Common difference (d) = a_{2} - a

= 80 - 84 = -4

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

Given nth term is 0

0 = 84 + (n - 1)× -4

⇒ 0 = 84 -4(n - 1)

⇒ 4(n-1) = 84

⇒ n - 1= 84/4 = 21

⇒ n = 21 + 1= 22

⇒ 22 nd term is 0.

(iii) Given A.P. 4, 9, 14, .......

First term (a) = 4

Common difference (d) = a_{2} - a

= 9 - 4 = 5

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

Given nth term is 254

4 + (n - 1)5 = 254

⇒ (n - 1)5 = 250

⇒ n - 1= 250/5 = 50

= 50 + 1 = 51

∴ 51 th term is 254.

(iv) Given A.P 21, 42, 63, 84, ......

a = 21, d = a_{2} - a

= 42 - 21 = 21

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

Given nth term = 420

21 + (n - 1)21 = 420

⇒ (n - 1)21 = 399

⇒ n - 1 = 399/21 = 19

⇒ n = 19 + 1 = 20

∴ 20th term is 420.

(v) Given A.P is 121, 117, 113,........

First term (a) = 121

Common difference (d) = 117 - 121 = -4

nth term (a) = a + (n - 1)d

Given nth term is negative i.e., a_{n} < 0

121 + (n - 1)-4 < 0

⇒ 121+ 4 - 4n < 0

⇒ 125 < 4n

⇒ n > 125/4

⇒ n > 31.25

The integer which comes after 31.25 is 32,

∴ 32 nd term is first negative term.

**3. (i) Is 68 a term of the AP 7, 10, 13, ...... ?(ii) Is 302 a term of the AP 3, 8, 13,...... ? (iii) Is -150 a term of the AP 11, 8, 5, 2, ...... ? **

**Solution**

In the given problem, we are given an A.P and the value of one of its term

We need to find whether it is a term of the AP or not so here we will use the formula

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

(i) Here, AP is 7, 10, 13, ......

a_{n} = 68, a = 7, and d = 10 - 7 = 3

Using the above mentioned formula, we get

68 = 7 + (n - 1)3

⇒ 68 - 7 = 3n - 3

⇒ 31 + 3 = 3n

⇒ 64 = 3n

⇒ n = 64/3

Since, the value of n is a fraction.

Thus, 68 is not the team of the given A.P.

(ii) Here, AP is 3, 8, 13,....

a_{n} = 302, a = 3

d = 8 - 3= 5 using the above mentioned formula, we get

302 = 3 + (n -1)5

⇒ 302 - 3 = 5n - 5

⇒ 299 = 5n - 5

⇒ 5n = 304

⇒ n = 304/5

Since, the value of 'n' is a fraction. Thus, 302 is not the term of the given A.P.

(iii) Here, A.P. is 11, 8, 5, 2, ......

a_{n} = -150, a = 1 and d = 8 - 11= -3

Thus, using the above mentioned formula, we get

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

⇒ -150 - 11 = -3n + 3

⇒ -161 - 3 = -3n

⇒ n = -164/-3

Since, the value of n is a fraction. Thus, -150 is not the term of the given A.P.

**4. How many term are there in the AP ? (i) 7, 10, 13,..... 43(ii)-1, -5/6, -2/3, -1/2,....., 10/3(iii) 7, 13, 19,...... 05(iv) 18, 15(1/2), 13, ...... , -47 **

**Solution**

(i) 7, 10, 13,..... 43

From given A.P

a = 7, d = 10 - 7 = 3, a_{n} = a + (n - 1)d.

Let, a_{n} = 43 **(last term)**

7 + (n - 1)3 = 43

⇒ (n-1) = 39/3=13

⇒ n = 13

∴ 13 terms are there in given A.P.

(ii) -1, -5/6, -2/3, -1/2,........, 10/3

From given A.P

a = -1, d = -5/6 + 1 = 1/6

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

Let, a_{n} = 10/3 **(last term)**

-1 + (n - 1)1/6 = 10/3

⇒ (n - 1) × 1/6 = 13/3

⇒ n - 1 = (13 × 6)/3 = 26

⇒ n = 27

∴ 27 terms are there in given A.P.

(iii) 7, 13, 19, ....., 05

From the given A.P

a = 7, d = 13 - 7 = 6,

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

Let a_{n} = 205 **(last term)**

7 + (n - 1)6 = 205

⇒ (n - 1)6 = 205- 7 = 198

⇒ n -1 = 198/6 = 33

⇒ n = 33+ 1 = 34

∴ 34 terms are there in given A.P.

(iv) 18, 15(1/2), 13, ...... , -47

From the given A.P

a = 18, d = 31/2 - 18 = 15.5 - 18 = -2.5

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

Let a_{n} = -47 **(last term)**

18 + (n - 1) × -2.5 = - 47

⇒ (n -1)× -2.5 = -47 - 18

⇒ n - 1= -65/-2.5

⇒ n - 1 = (65 × 10)/25 = 26

⇒ n = 26 + 1 = 27

∴ 27 terms are there in given A.P.

**5. The first term of an AP is 5, the common difference is 3 and the last term is 80, find the number of terms. **

**Solution**

Given,

First term(a) = 5

Common difference (d) = 3

Last term (a_{n} ) = 80

To calculate no of terms in given A.P

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

Let a_{n} = 80,

80 = 5 + (n - 1)3

⇒ (n - 1)3 = 75

⇒ n - 1 = 75/3 = 25

⇒ n = 25 + 1 = 26

∴ There are 26 terms.

**6. The 6th and 17 terms of an A.P are 19 and 41 respectively, find the 40th term. **

**Solution**

Given, a_{6} = 19, a_{17} = 41

⇒ a_{6} = a + (6 - 1)d

19 = a + 5d **...(1)**

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

⇒ 41 = a + 16d **...2) **

Subtract (1) from (2)

a + 16d = 41 **...(1)**

a + 5d = 19 **...(2)**

11d = 22

⇒ d = 22/11 = 2

Substitute d = 2 in (1)

19 = a + 5(2)

a = 9

∴ 40th term a_{40} = a + (40 - 1)d

= 9 + 39×2

= 9 + 78 = 87

∴ a_{40} = 87

**7. If 9th term of an A.P is zero, prove that its 29th term is double the 19th term. **

**Solution**

Given,

9th term of an A.P a_{9} = 0,

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

a + (a - 1)d = 0

⇒ a + 8d = 0

⇒ a = -8d

We have to prove,

24th term is double the 19th term a_{29} = 2. a_{19}

a + (29 - 1)d = 2[a + (19 - 1)d]

a + 28d = 2[a + 18d]

Put a = -8d

-8d + 28d = 2[-8d + 18d]

⇒ 20d = 2×10d

⇒ 20d = 20d

Hence, proved.

**8. If 10 times the 10th term of an A.P is equal to 15 times the 15th term, show that 25th term of the A.P. is zero. **

**Solution**

Given,

10 times of 10th term is equal to 15 times of 15th term.

10a_{10} = 15.a_{15 }

10[a + (10 - 1)d] = 15[a + (15 - 1)d]** [∴ a _{n} = a + (n - 1)d]**

⇒ 10(a + 9d) = 15(a + 14.d)

⇒ a + 9d = 15/10(a + 14d)

⇒ a - (3/2)a = 42d/2 - 9d

⇒ -1a/2 = (42d - 18d)/2

⇒ -a/2 = 24d/2 =12d

⇒ a = -24d

We have to prove 25 th term of A.P is 0

a

_{25}= 0

⇒ a + (25 - 1)d = 0

⇒ a + 24d = 0

Put a = -24d

-24d + 24d = 0

⇒ 0 = 0

Hence proved.

**9. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term. **

**Solution**

Given,

a_{n} = 41, a_{n} = 73,

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

⇒ a_{10} = a + (10 - 1)d

41 = a + 9d** ...(1) **

⇒ a_{n} = a + (18 - 1)d

73 = a + 17d **...(2)**

Subtract (1) from (2)

a + 17d = 73 **...(1) **

⇒ a + 9d = 41 **...(2)**

⇒ 0 + 8d = 32

⇒ d = 32/8 = 4

Substitute d = 4 in (1)

a + 9×4 = 41

⇒ a = 41 - 36

⇒ a = 5

26th term a_{n} = a + (26 - 1)d

= 5 + 25×4

= 5 + 100

= 105

∴ 26th term a_{26} = 105.

**10. In a certain A.P the 24th term is twice the 10th term. Prove that the 72 nd term is twice the 34th term. **

**Solution**

Given,

24th term is twice the 10th term

a_{24} = 2 a_{10}

Let, first term of a square = a

Common difference = d

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

a + (24 - 1)d = (a + (10 - 1)d

⇒ a + 23d = 2(a + 9d)

⇒ (23 - 18)d = a

⇒ a = 5d

We have to prove,

72nd term is twice the 34th term

a_{12} = 2a_{34}

⇒ a + (12 - 1)d = 2[a + (34 - 1)d]

⇒ a + 71d = 2a + 66d

Substitute a = 5d

5d + 71d = 2(5d) + 66d

⇒ 76d = 10 d + 66d

⇒ 76d = 76d

Hence proved.

**11. If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n +1)th term. **

**Solution**

Given,

(m + 1) th term is twice the (m + 1)th term.

First term = a

Common difference = d

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

a_{m+1} = 2a_{n} + 1

⇒ a + (m + 1 - 1)d = 2(a + (n + 1 - 1)d

⇒ a + md = 2(a + nd)

⇒ a = (m - 2n)d

We have to prove

(3m + 1)th term is twice the (m + n + 1)th term

a_{3m + 1} = 2.a_{m+n+1}

⇒ a + (3m + 1 - 1)d = (a + (m + n + 1 -1)d

⇒ a + 3m.d - 2a + 2(m + n)d

Substitute a = (m - 2n)d

(m - 2n)d + 3m d = 2(m - 2n)d + 2(m + n)d

⇒ 4m - 2n = 4m - 4n + 2n

⇒ 4m - 2n = 4m - 2n

Hence, proved.

**12. If the n term of the A.P 9, 7, 5, ...... is same as the term of the A.P. 15, 12, 9, ..... find n. **

**Solution**

Given,

First sequence is 9,7,5, .....

a = 9, d = 7 - 9 = -2, a_{n} = a + (n + 1)d

a_{n} = 9 + (n - 1)×-2

Second sequence is 15, 12, 9, .....

a = 15, d = 12 - 15 = -3, a_{n} = a + (n - 1)d

a_{n} = 15 + (n - 1) × -3

Given an a_{n} are equal

9 -2(n - 1) = 15 - 3(n - 1)

⇒ 3(n - 1) - 2(n - 1) - 15 - 9

⇒ n - 1 = 6

⇒ n = 7

∴ 7th term of two sequence are equal.

**13. Find the 12th term from the end of the following arithmetic progressions: (i) 3, 5, 7, 9, .....201 (ii) 3, 8,13, ......, 253(iii) 1, 4,7, 10, ...... 88**

**Solution**

(i) 3, 5, 7, 9, ....., 201

First term (a) = 3

Common difference (d) = 5 - 3 = 2

12th term from the end is can be considered as (l) last term = first term and common difference = d_{1} = -d

nth term from the end = last term + (n - 1)×-d

12 th term from end = 201 + (12 - 1)× -2

= 201 - 22

= 179

(ii) 3, 8, 13, .....,253

First term a = 3

Common difference d = 8 - 3 = 5

Last term (l) = 253

n th term of sequence on = a + (n - 1)d

To find nth term from the end, we put last term (l) as 'a' and common difference as -d

nth term from the end = last term + (n - 1) -d

12th term from the end = 253 + (12 - 1) ×-5

= 253 - 55 = 198

∴ 12 th term from the end = 198

(iii) 1, 4, 7, 10, ......., 88

First term a = 1

Common difference d = 4 - 1 = 3

Last term (l) = 88

n th term a_{n} = a + (n + 1)d

n th term from the end = last term + (n - 1)-d

12 th term from the end = 88 + (12 - 1)× -3

= 88 - 33

= 55

∴ 12th term from the end = 55

**14. The 4th term of an A.P is three times the first and the 7 term exceeds twice the third term by 1. Find the first term and the common difference. **

**Solution**

Given,

4 th term of an AP is three times the times the first term

a_{4} = 3.a

n th term of a sequence a_{n} = a + (n - 1)d

a + (4 - 1)d = 3a

⇒ a + 3d = 3a

⇒ 3d = 2a

⇒ a = 3d/2 ** ...(1) **

Seventh term exceeds twice the third term by 1.

a_{7} + 1 = 2a_{3}

⇒ a + (7 - 1)d + 1 = 2[a+(3 -1)d]

⇒ a + 6d + 1 = 2a + 4d

⇒ 2a - a = 6d - 4d + 1

⇒ a = 2d + 1 ** ...(2) **

By equating (1), (2)

3d/2 = 2d + 1

⇒ 3d/2 - 2d = 1

⇒ -d/2 = 1

⇒ d = -2

Put d = -2 in a = 3d/2

= (3/2)(-2)

= -3

∴ First term a = -3, common difference d = -2.

**15. Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22. **

**Solution**

Given,

a_{6} = 12, a_{8} = 22

nth term of an A.P a_{n} = a + (n - 1)d

a_{6} = a + (6 - 1)d = a + 5d = 12 **...(1) **

a_{8} = a+ (8 -1)d = a + 7d = 22 ** ...(2) **

Subtracting (1) from (2)

2d = 10

⇒ d = 5

putting value of d in equation (1)

a + 5d = 12

⇒ a + 5×5 = 12

⇒ a = 12 - 25 = -13

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

=a + d

= -13 + 5 = -8

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

= -13 + (n - 1)×5

= -13 + 5n - 5

= -18 + 5n

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

= -13 + (n - 1)-5

a_{n} = -18 + 5n

∴ a_{2} = -8, a_{n} = -18 + 5n

**16. How many number of two digits are divisible by 3 ? **

**Solution**

We observe that 12 is the firs two - digit number divisible by 3 and 99 is the last two digit number divisible by 3. Thus, the sequence is

12, 15, 18,..., 99

This sequence is in A.P with

First term (a) = 12

Common difference(d) = 15 - 12 = 3

nth term a_{n} = 99

nth term of an A.P

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

99 = 12 +(n -1)3

⇒ 99 - 12 = (n - 1)3

⇒ 87/3 = n = 1

⇒ n = 30

∴ 30 term are there in the sequence.

**17. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32 nd term. **

**Solution**

Given,

No. of terms = n = 60

First term (a) = 7

Last term (a) = 7

Last term (a_{10}) = 125

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

⇒ a_{60} = a + (60 - 1)d

⇒ 125 = 7 + 59d

⇒ 118 = 59d

⇒ d = 118/59 = 2

52 nd term a_{32} = a + (32 - 1)d

= 7 + 31×2

= 7 + 62 = 69

**18. The sum of 4 and 8th terms of an A.P. is 24 and the sum of the 6th and 10 th terms is 34. Find the first term and the common difference of the A.P. **

**Solution**

Given,

a_{4} + a_{8} = 24

a_{6} + a_{10} = 34

⇒ a + (4 - 1)d + a + (18 - 1)d = 24

2a + 10d = 24

a + 5d = 12 **....(1)**

⇒ a_{6} + a_{10} = 34

a + (6 - 1)d + a + (10 - 1)d = 34

⇒ 2a + 14d = 34

a + 7d = 17 **....(2) **

Subtract (1) from (2)

a + 7d = 17

⇒ a + 5d = 12

⇒ 2d = 5

⇒ d = 5/2

Put d = 5/2 in a + 5d = 12

a = 12 - 5d

⇒ a = 12 - 5× 5/2

⇒ a = 12 - 25/2 = -1/2

∴ a = -1/2, d = b/2

**19. The first term of an A.P. is 5 and its 100th term is -292. Find the 50th term of this A.P. **

**Solution**

Given,

First term of an A.P. = 5 and 100th term = -292

**20. Find a _{30} – a_{20} for the A.P.**

(i) -9, -14, -19, -24, …

(ii) a, a + d, a + 2d, a + 3d, …

**Solution**

**21. Write the expression a _{n} – a_{k} for the A.P. a, a + d, a + 2d, ……**

Hence, find the common difference of the A.P. for which

(i) 11th term is 5 and 13th term is 79.

(ii) a_{10} – a_{5} = 200

(iii) 20th term is 10 more than the 18th term.

**Solution**

In the A.P. a, a + d, a + 2d, ….

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

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

= a + nd - d - (a + kd - d)

= a + nd - d - a - kd + d

= nd - kd - (n - k)d

**22. Find n if the given value of x is the nth term of the given A.P.**

**(i) 25, 50, 75, 100,....; x = 1000(ii) -1, -3, -5, -7, ....; x = -151(iii) 5½, 11, 16½, 22, ....; x = 550(iv) 1, 21/11, 31/11, 41/11, ....; x = 171/11**

**Solution**

(i) The A.P. is 25, 50, 75, 100, ...; x = 1000

Here, a = 25 and d = 50 - 25 = 25 and a_{n} = 100

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

1000= 25 + (13-1)×25

= 25 + 25n - 25 = 25n

(ii) The A.P. is -1, -3, -5, -7, ...; x = -151

Here, a = -1, d = -3 -(-1) = -3 + 1 = -2 and a_{n} = -15

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

⇒ -151 = -1 + (n-1)(-2)

⇒ -151 = -1 - 2n + 2

⇒ -151 = 1 - 2n

⇒ -2n = -151 - 1 = -152

**23. The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.**

**Solution**

**24. Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.**

**Solution**

Let a, a + d, a + 2d, a + 3d, …. be the A.P.

a_{n} = a + (n – 1) d

But a_{3} = 16

**25. The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P.**

**Solution**

Let a, a + d, a + 2d, a + 3d, be the A.P.

Here a is the first term and d is the common difference

a_{n} = a + (n – 1) d

Now, a_{7} = a + (7 – 1) d = a + 6d = 32 **….(i)**

and a_{13} = a + (13 – 1) d = a + 12d = 62 **….(ii)**

Subtracting (i) from (ii)

6d = 30

⇒ d = 5

a + 6×5 = 32

⇒ a + 30 = 32

⇒ a = 32 – 30 = 2

A.P. will be 2, 7, 12, 17, ……

**26. Which term of the A.P. 3, 10, 17, … will be 84 more than its 13th term ?**

**Solution**

The given A.P. is 3, 10, 17,... whose first term (a) = 3 and,

common difference (d) = 10 - 3 = 7

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

Let kth term is greater than 13th term by 84.

kth term(a_{k}) = a + (k-1)d

and a_{13} = a + (13-1) = a + 12d

But a_{k} - a_{13} = 84

a + (k-1)d - (a + 12d) = 84

⇒ a + kd - d - a - 12d = 84

⇒ kd - 13d = 84

⇒ (k-13)d = 84

⇒ 7(k-13) = 84

⇒ 7k - 91 = 84

**27. Two arithmetic progressions have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms ? **

**Solution**

**28. For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ...... and 3, 10, 17, ..... are equal ? **

**Solution**

63, 65, 67,.... and 3, 10, .....

a = 63 and d = 65 – 63 = 2

a

_{n}= a

_{1}+ (n–1) d = 63 + (n–1)×2 = 63 + 2n – 2 = 61 + 2n

and in the A.P. 3, 10, 17, …

a = 3 and d = 10 – 3 = 7

a

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

But both nth terms are equal

61 + 2n = 7n – 4

⇒ 61 + 4 = 7n – 2n

⇒ 65 = 5n

⇒ n = 13

**29. How many multiples of 4 lie between 10 and 250 ?**

**Solution**

All the terms between 10 and 250 are multiple of 4

First multiple (a) = 12

and last multiple (l) = 248

and d = 4

Let n be the number of multiples, then

a_{n} = a + (n–1) d

⇒ 248 = 12 + (n–1)×4 = 12 + 4n – 4

⇒ 248 = 8 + 4n

⇒ 4n = 248 – 8 = 240

n = 60

Number of terms are = 60

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

**Solution**

First three digit number is 100 and last three digit number is 999

In the sequence of the required three digit numbers which are divisible by 7, will be between

a = 105 and last number l = 994 and d = 7

Let n be the number of terms, then

a_{n} = a + (n – 1) d

994 = 105 + (n–1)×7

⇒ 994 = 105 + 7n – 7

⇒ 7n = 994 – 105 + 7

⇒ 7n = 896

⇒ n = 128

Number of terms =128

**31. Which term of the arithmetic progression 8, 14, 20, 26, … will be 72 more than its 41st term ? **

**Solution**

a

_{n}= a

_{41}+ 72

For the given sequence

a = 8, d = 14 - 8 = 6

⇒ a + (n - 1)d = 8 + (n -1)6 + 72

⇒ 8 + (n - 1)6 = 8 + 40×6 + 72

⇒ (n - 1)6 = 312

⇒ n - 1 = 312/6 = 52

⇒ n = 52+1 = 53

∴ 53rd term is 72 more than 41st term

**32. Find the term of the arithmetic progression 9, 12, 15, 18, … which is 39 more than its 36th term.**

**Solution**

In the given A.R 9, 12, 15, 18, …

First term (a) = 9

and common difference (d) = 12 – 9 = 3

and a_{n} = a + (n – 1) d

Now a_{36} = a + (36 – 1) d = 9 + 35×3 = 9 + 105 = 114

Let the an be the required term

a_{n} = a + (n – 1) d

= 9 + (n – 1)×3 = 9 + 3n – 3 = 6 + 3n

But their difference is 39

a_{n} – a_{36} = 39

⇒ 6 + 3n – 114 = 39

⇒ 114 – 6 + 39 = 3n

⇒ 3n = 147

⇒ n = 49

Required term is 49th.

**33. Find the 8th term from the end of the A.P. 7, 10, 13, …, 184.**

**Solution**

The given A.P. is 7, 10, 13,…, 184

Here first term (a) = 7

and common difference (d) = 10 – 7 = 3

and last tern (l) = 184

Let nth term from the last is a_{n} = l – (n–1) d

a_{8}= 184 – (8–1)×3 = 184 – 7×3 = 184 – 21 = 163

**34. Find the 10th term from the end of the A.P. 8, 10, 12, …, 126.**

**Solution**

The given A.P. is 8, 10, 12, …, 126

Here first term (a) = 8

Common difference (d) = 10 – 8 = 2

and last tern (l) = 126

Now nth term from the last is a_{n} = l – (n–1) d

a_{10} = 126 – (10–1)×2

= 126 – 9×2

= 126 – 18

= 108

**35. The sum of 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the A.P.**

**Solution**

**36. Which term of the A.P. 3, 15, 27, 39, ..... will be 120 more than its 21st term?**

**Solution**

Here first term (a) = 3 and c.d. (d) = 15 - 3 = 12

Let nth term be the required term

Now,

21st term = a + (n – 1) d = 3 + 20 x 12 = 3 + 240 = 243

According to the given condition,

nth term – 21 st term = 120

⇒ a + (n – 1) d – 243 = 120

⇒ 3 + (n – 1)×12 = 120 + 243 = 363

⇒ (n – 1) 12 = 363 – 3 = 360

⇒ n – 1 = 30

⇒ n = 30 + 1 = 31

31st term is 120 more than 21st term.

**37. The 17th term of an A.P is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the nth term. **

**Solution**

**38. Find the number of ail three digit natural numbers which are divisible by 9.**

**Solution**

d= 9

a + (n – 1) d = 999

⇒ 108 + (n – 1)×9 = 999

⇒ (n – 1) d = 999 – 108

⇒ (n – 1)×9 = 891

⇒ n – 1 = 99

⇒ n = 99 + 1 = 100

Number of terms = 100

**39. The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.**

**Solution**

**40. The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.**

**Solution**

Let a be the first term and d be the common difference and

T_{n} = a + (n–1) d

**41. The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term.**

**Solution**

Let a be the first term, d be the common difference, then

T_{n} = a + (n–1) d

Hence, 72nd term = 4 times of 15th term.

**42. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.**

**Solution**

Numbers divisible by both 2 and 5 are 110, 120, 130, ………. , 990

Here a = 110, x = 120 – 110 = 10

a_{n} = 990

A,

a + (n – 1) d = 990

⇒ 110 + (n – 1) (10) = 990

⇒ (n – 1) (10) = 990 – 110 = 880

⇒ n – 1 = 88

⇒ n = 88 + 1 = 89

**43. If the seventh term of an AP is 1/9 and its ninth term is 1/7, find its 63rd term.**

**Solution**

**44. The sum of 5th and 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP.**

**Solution**

**45. Find where 0 (zero) is a term of the AP 40, 37, 34, 31, ……**

**Solution**

AP 40, 37, 34, 31, …..

Here, a = 40, d = -3

Let T_{n} = 0

T_{n} = a + (n – 1) d

⇒ 0 = 40 + (n – 1) (-3)

⇒ 0 = 40 – 3n + 3

⇒ 3n = 43

⇒ n = 433 which is in fraction.

There is no term which is 0.

**46. Find the middle term of the A.P. 213, 205, 197, …, 37.**

**Solution**

= 213 - 88 = 125

∴ Middle term = 125.

**47. If the 5th term of an A.P. is 31 and 25th term is 140 more than the 5th term, find the A.P.**

**Solution**

We know that,

T_{n} = a + (n – 1 )d

T_{5} = a + 4d

⇒ a + 4d = 31 **…(i)**

and T_{25} = a + 24d

⇒ a + 24d = 140 + T_{5}

⇒ a + 24d = 140 + 31 = 171 **…(ii)**

Subtracting (i) from (ii),

20d= 140

and a + 4d = 31

⇒ a + 4×7 = 31

⇒ a + 28 = 31

⇒ a = 31 – 28 = 3

a = 3 and d = 7

AP will be 3, 10, 17, 24, 31, ….

**48. Find the sum of two middle terms of the A.P.-4/3, -1, -2/3, -1/3,...., 4 1/3**

**Solution**

**49. If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m+n+1)th term.**

**Solution**

Let a, a+d, a+2d, a+3d, .... is in A.P.

∴ (m+1)th term = a + (m+1-1)d

= a + md

and,

(n+1)th term = a + (n+1-1)d

= a + nd

∵ (m+1)th term = 2(n+1)th term

∴ a + md = 2(a + nd)

⇒ a + md = 2a + 2nd

⇒ 2a - a = md - 2 nd

⇒ a = d(m - 2n) = (m - 2n) d** ...(i)**

Now,

(3m + 1)th term = a+(3m + 1 - 1) d

= (m - 2n) d + 3md

= (m - 2n + 3m)d

= (4m - 2n)d

= 2 (2m - n) d** ...(ii)**

and (m+n+1)th term = a+(m+n+1-1)d

= a+(m+n)d

= (m-2n)d + (m+n)d

= (m-2n+m+n)d

= (2m - n)d **...(iii)**

From (ii) and (iii),

(3m + 1)th term = 2 (m + n + 1)th term.

Hence, proved.

**50. If an A.P. consists of n terms with first term a and nth term l show that the sum of the mth term from the beginning and the mth term from the end is (a + l).**

**Solution**

In an A.P.,

Number of terms = n

First term = a

and nth term = l

mth term (a_{m}) = a + (m – 1) d

and mth term from the end = l – (m – 1)d

Their sum = a + (m – 1) d + l – (m – 1) d = a + l

Hence proved.

**51. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?**

**Solution**

Here, the first number is 11, which divided by 4 leave remainder 3 between 10 and 300.

Last term before 300 is 299, which divided by 4 leave remainder 3.

11, 15, 19, 23, …, 299

Here, first term (a) = 11,

common difference (d) = 15 – 11 = 4

nth term, a_{n} = a + (n – 1 ) d = l **[last term]**

⇒ 299 = 11 + (n – 1) 4

⇒ 299 – 11 = (n – 1) 4

⇒ 4(n – 1) = 288

⇒ (n – 1) = 72

n = 73

**52. Find the 12th term from the end of the A.P. -2, -4, -6, …, -100.**

**Solution**

Given, A.P., -2, -4, -6, …, -100

Here, first term (a) = -2,

common difference (d) = -4 - (-2)

and the last term (l) = -100.

We know that, the nth term an of an A.P. from the end is a_{n} = l - (n - 1 )d,

where l is the last term and d is the common difference. 12th term from the end,

a_{n} = -100 - (12 - 1) (-2)

= -100 + (11) (2)

= -100 + 22

= -78

Hence, the 12th term from the end is -78.

**53. For the A.P.: -3, -7, -11,…, can we find a _{30} – a_{20} without actually finding a_{30} and a_{20} ? Give reasons for your answer.**

**Solution**

True.

nth term of an A.P., a_{n} = a + (n – 1)d

a_{30} = a + (30 – 1 )d = a + 29d

and a_{20} = a + (20 – 1 )d = a + 19d **…(i)**

Now, a_{30} – a_{20} = (a + 29d) – (a + 19d) = 10d

and from given A.P.

common difference, d = -7 – (-3) = -7 + 3 = -4

a_{30} – a_{20 }= 10(-4) = -40 **[from Eq- (i)]**

**54. Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?**

**Solution**

Let the same common difference of two A.P.’s is d.

Given that, the first term of first A.P. and second A.P. are 2 and 7 respectively,

then the A.P.’s are 2, 2 + d, 2 + 2d, 2 + 3d, … and 7, 7 + d, 7 + 2d, 7 + 3d, …

Now, 10th terms of first and second A.P.’s are 2 + 9d and 7 + 9d, respectively.

So, their difference is 7 + 9d – (2 + 9d) = 5

Also, 21st terms of first and second A.P.’s are 2 + 20d and 7 + 20d, respectively.

So, their difference is 7 + 20d – (2 + 9d) = 5

Also, if the a_{n} and b_{n} are the nth terms of first and second A.P.

Then b_{n} - a_{n} = [7 + (n - 1) d] – [2 + (n - 1) d = 5

Hence, the difference between any two corresponding terms of such A.P.’s is the same as the difference between their first terms.