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Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progressions – Exercise 9.4 | Set 1
  • Last Updated : 25 Jan, 2021

Question 1. Find:

(i) 10th tent of the AP 1, 4, 7, 10….

(ii) 18th term of the AP √2, 3√2, 5√2, …….

(iii) nth term of the AP 13, 8, 3, -2, ……….

(iv) 10th term of the AP -40, -15, 10, 35, ………….

(v) 8th term of the AP 11, 104, 91, 78, ……………



(vi) 11th tenor of the AP 10.0, 10.5, 11.0, 11.2, …………..

(vii) 9th term of the AP 3/4, 5/4, 7/4 + 9/4, ………..

Solution: 

(i) Given that,

A.P. is 1, 4, 7, 10, ……….

First term(a) = 1

Common difference (d) = Second term – First term

= 4 – 1 = 3.

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 10th term in the A.P is 1 + (10 – 1)3

= 1 + 9×3 = 1 + 27 = 28

Hence 10th term of A.P is 28

(ii) Given that,

A.P. is √2, 3√2, 5√2, …….

First term (a) = √2,

Common difference = Second term – First term

d = 3√2 – √2 = 2√2

As we know that, to find nth term in an A.P = a + (n – 1)d



Therefore, 18th term of A. P. = √2 + (18 – 1)2√2

= √2 + 17.2√2 = √2 (1+34) = 35√2

Hence 18th term of A.P is 35√2

(iii) Given that,

A. P. is 13, 8, 3, – 2, …………

First term (a) = 13,

Common difference (d) = Second term first term

d = 8 – 13 = – 5

As we know that, to find nth term in an A.P = a + (n – 1)d

= 13 + (n – 1) – 5 = 13 – 5n + 5

Hence nth term of the A.P is an = 18 – 5n

(iv) Given that,

A. P. is – 40, -15, 10, 35, ……….

First term (a) = -40,

Common difference (d) = Second term – fast term

d = -15 – (- 40) = 40 – 15 = 25

As we know that, to find nth term in an A.P = a + (n – 1)d

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

= – 40 + 9.25 = – 40 + 225 = 185

Hence 10th term of the A.P is 185



(v) Given that,

Sequence is 117, 104, 91, 78, ………….

First term (a) = 117,

Common difference (d) = Second term – first term

d = 104 – 117 = –13

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 8th term = a + (8 – 1)d

= 117 + 7(-13) = 117 – 91 = 26

Hence 8th term of the A.P is 26

(vi) Given that,

A. P is 10.0, 10.5, 11.0, 11.5,

First term(a) = 10.0,

Common difference (d) = Second term – first term

d = 10.5 – 10.0 = 0.5

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 11th term a11 = 10.0 + (11 – 1)0.5

= 10.0 + 10 x 0.5 = 10.0 + 5 =15.0

Hence 11th term of the A. P. is 15.0

(vii) Given that,

A. P is 3/4, 5/4, 7/4, 9/4, …………

First term (a) = 3/4,

Common difference (d) = Second term – first term

d = 5/4 – 3/4 = 2/4

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore, 9th term a9 = a + (9 – 1)d

= 3/4 + 16/4 = 19/4

Hence 9th term of the A.P is 19/4.

Question 2. Find:

(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 that,

A.P. is 3, 8, 13, ………..

First term (a) = 3,

nth term is 248

Common difference (d) = Second term – first term

d = 8 – 3 = 5

As we know that, to find nth term in an A.P = a + (n – 1)d

248 = 3+(n – 1)5

248 = -2 + 5n

5n = 250

n =250/5 = 50

Hence 50th term in the A.P is 248.

(ii) Given that,

A. P is 84, 80, 76, …………

First term (a) = 84

nth term is 0

Common difference (d) = a2 – a

d = 80 – 84 = – 4

As we know that, to find nth term in an A.P = a + (n – 1)d

0 = 84 + (n – 1) – 4

84 = +4(n – 1)

n – 1 = 84/4 = 21

n = 21 + 1 = 22

Hence 22nd term in the A.P is 0.

(iii) Given A. P 4, 9, 14, …………

First term (a) = 4,

nth term is 254

Common difference (d) = a2 – a1

d = 9 – 4 = 5

As we know that, to find nth term in an A.P = a + (n – 1)d

4 + (n – 1)5 = 254

(n – 1)∙5 = 250

n – 1 = 250/5 = 50

n = 51

Hence 51th term in the A.P is 254.

(iv) Given that,

A. P 21, 42, 63, 84, ………

a = 21,

nth term = 420,

d = a2 – a1

= 42 – 21 = 21

As we know that, to find nth term in an A.P = a + (n – 1)d

21 + (n – 1)21 = 420

(n – 1)21 = 399

n – 1 = 399/21 = 19

n = 20

Hence 20th term is 420.



(v) Given that,

A.P is 121, 117, 113, ………..

First term (a) = 121,

nth term is negative i.e. an < 0,

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

As we know that, to find nth term in an A.P = a + (n – 1)d

121 + (n – 1) – 4 < 0

121 + 4 – 4n < 0

125 – 4n < 0

4n > 125

n > 125/4

n > 31.25

The integer which comes after 31.25 is 32.

Hence 32nd term in the A.P will be the first negative term.

Question 3. 

(i) Is 68 a term of the A.P. 7, 10, 13,… ?

(ii) Is 302 a term of the A.P. 3, 8, 13, …. ?

(iii) Is -150 a term of the A.P. 11, 8, 5, 2, … ?

Solutions: 

(i) Given that,

A.P. 7, 10, 13,…

from given series,

a = 7 and d = a2 – a1 = 10 – 7 = 3

As we know that, to find nth term in an A.P = a + (n – 1)d

we have to find at which position 68 is present in given series,

a + (n – 1)d = 68

7 + (n – 1)3 = 68

7 + 3n – 3 = 68

3n + 4 = 68

3n = 64

n = 64/3, which is not a whole number.

Hence, 68 is not a term in the A.P.

(ii) Given, A.P. 3, 8, 13,…

from given series, a = 3 and d = a2 – a1 = 8 – 3 = 5

As we know that, to find nth term in an A.P = a + (n – 1)d

we have to find at which position 302 is present in given series,

a + (n – 1)d = 302

3 + (n – 1)5 = 302

3 + 5n – 5 = 302

5n – 2 = 302

5n = 304

n = 304/5, which is not a whole number.

Hence, 302 is not a term in the A.P.

(iii) Given, A.P. 11, 8, 5, 2, …

from given series,

a = 11 and d = a2 – a1 = 8 – 11 = -3

As we know that, to find nth term in an A.P = a + (n – 1)d

we have to find at which position -150 is present in given series,

a + (n – 1)d = -150

11 + (n – 1)(-3) = -150

11 – 3n + 3 = -150

3n = 150 + 14

3n = 164

n = 164/3, which is not a whole number.



Hence, -150 is not a term in the A.P.

Question 4. How many terms are there in the A.P.?

(i) 7, 10, 13, ….., 43

(ii) -1, -5/6, -2/3, -1/2, … , 10/3

(iii) 7, 13, 19, …, 205

(iv) 18, 15½, 13, …., -47

Solution: 

(i) Given that,

A.P. 7, 10, 13, ….., 43

where, a = 7 and d = a2 – a1 = 10 – 7 = 3

As we know that, to find nth term in an A.P = a + (n – 1)d

a + (n – 1)d = 43

7 + (n – 1)(3) = 43

7 + 3n – 3 = 43

3n = 43 – 4

3n = 39

n = 13

Hence, there are 13 terms in the given A.P.

(ii) Given that,

A.P. -1, -5/6, -2/3, -1/2, … , 10/3

where, a = -1 and d = a2 – a1 = -5/6 – (-1) = 1/6

As we know that, to find nth term in an A.P = a + (n – 1)d

a + (n – 1)d = 10/3

-1 + (n – 1)(1/6) = 10/3

-1 + n/6 – 1/6 = 10/3

n/6 = 10/3 + 1 + 1/6

n/6 = (20 + 6 + 1)/6

n = (20 + 6 + 1)

n = 27

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

(iii) Given that,

A.P. 7, 13, 19, …, 205

where, a = 7 and d = a2 – a1 = 13 – 7 = 6,

nth term is 205

As we know that, to find nth term in an A.P = a + (n – 1)d

a + (n – 1)d = 205

7 + (n – 1)(6) = 205

7 + 6n – 6 = 205

6n = 205 – 1

n = 204/6

n = 34

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

(iv) Given that,

A.P. 18, 15½, 13, …., -47

where, a = 7 and d = 15½ – 18 = 5/2,

As we know that, to find nth term in an A.P = a + (n – 1)d

a + (n – 1)d = 43

18 + (n – 1)(-5/2) = -47

18 – 5n/2 + 5/2 = -47

36 – 5n + 5 = -94

5n = 94 + 36 + 5

5n = 135

n = 27

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

Question 5. The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms.

Solution: 

Given that,

a = 5 and d = 3,

last term = 80

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore, for the given A.P. an = 5 + (n – 1)3 = 3n + 2

=3n + 2 = 80



3n = 78

n = 78/3 = 26

Hence, there are 26 terms in the A.P.

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

Solution: 

Given that,

a6 = 19 and a17 = 41

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore,

a6 = a + (6-1)d

= a + 5d = 19 ——-(i)

Similarly,

a17 = a + (17 – 1)d

= a + 16d = 41 ———-(ii)

Solving (i) and (ii),

(ii) – (i)

a + 16d – (a + 5d) = 41 – 19

11d = 22

d = 2

Using d in eqn(i), we get

a + 5(2) = 19

a = 19 – 10 = 9

Now, the 40th term is given by a40 = 9 + (40 – 1)2 = 9 + 78 = 87

Hence the 40th term is 87.

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

Solution: 

Given that,

a9 = 0

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore, a + (9 – 1)d = 0 ⇒ a + 8d = 0 ———-(i)

Now,

29th term is given by a29 = a + (29 – 1)d

=a29 = a + 28d

And, a29 = (a + 8d) + 20d (using (i))

= a29 = 20d ———-(ii)

Similarly, 19th term is given by a19 = a + (19 – 1)d

=a19 = a + 18d

And, a19 = (a + 8d) + 10d (using (i))

=a19 = 10d ———(iii)

On comparing (ii) and (iii), we observe that

a29 = 2(a19)

Hence, 29th term is double the 19th term.

Question 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 that,

10 times the 10th term of an A.P. is equal to 15 times the 15th term.

As we know that, to find nth term in an A.P = a + (n – 1)d

10(a10) = 15(a15)

10(a + (10 – 1)d) = 15(a + (15 – 1)d)

10(a + 9d) = 15(a + 14d)

10a + 90d = 15a + 210d

5a + 120d = 0

5(a + 24d) = 0

a + 24d = 0

a + (25 – 1)d = 0

a25 = 0

Hence, the 25th term of the A.P. is zero.

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

Solution: 

Given that,

A10 = 41 and a18 = 73

As we know that, to find nth term in an A.P = a + (n – 1)d

therefore,

a10 = a + (10 – 1)d

= a + 9d = 41 ———(i)

Similarly,

a18 = a + (18 – 1)d

= a + 17d = 73 ——-(ii)

Solving (i) and (ii),

(ii) – (i)

a + 17d – (a + 9d) = 73 – 41

8d = 32

d = 4

Using d in (i), we get



a + 9(4) = 41

a = 41 – 36 = 5

Now, the 26th term is given by a26 = 5 + (26 – 1)4 = 5 + 100 = 105

Hence the 26th term is 105.

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

Solution: 

Given that,

24th term is twice the 10th term.

As we know that, to find nth term in an A.P = a + (n – 1)d

a24 = 2(a10)

a + (24 – 1)d = 2(a + (10 – 1)d)

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a = 5d …. (1)

Now, the 72nd term can be expressed as:

a72 = a + (72 – 1)d

= a + 71d

= a + 5d + 66d

= a + a + 66d [using (1)]

= 2(a + 33d)

= 2(a + (34 – 1)d)

= 2(a34)

⇒ a72 = 2(a34)

Hence, the 72nd term is twice the 34th term of the given A.P.

Question 11. The 26th, 11th and the last term of an A.P. are 0, 3 and -1/5, respectively. Find the common difference and the number of terms.

Solution: 

Given that,

a26 = 0, a11 = 3 and an (last term) = -1/5 of an A.P.

As we know that, to find nth term in an A.P = a + (n – 1)d

Therefore,

a26 = a + (26 – 1)d

a + 25d = 0 ——–(1)

a11 = a + (11 – 1)d

a + 10d = 3 ———(2)

Solving (1) and (2),

(1) – (2)

a + 25d – (a + 10d) = 0 – 3

15d = -3

d = -1/5

Using d in (1), we get

a + 25(-1/5) = 0

a = 5

Now, given that the last term is -1/5

5 + (n – 1)(-1/5) = -1/5

5 + -n/5 + 1/5 = -1/5

25 – n + 1 = -1

n = 27

Hence, the A.P has 27 terms and its common difference is -1/5.

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

Solution: 

Given that,

A.P1 = 9, 7, 5, …. and A.P2 = 15, 12, 9, …

nth term of the A.P1 = nth term of the A.P2,

As we know that, to find nth term in an A.P = a + (n – 1)d

For A.P1,

a = 9, d = Second term – first term = 9 – 7 = -2

And, its nth term an = 9 + (n – 1)(-2) = 9 – 2n + 2

an = 11 – 2n ———-(i)

Similarly, for A.P2

a = 15, d = Second term – first term = 12 – 15 = -3

And, its nth term an = 15 + (n – 1)(-3) = 15 – 3n + 3

an = 18 – 3n ——–(ii)

11 – 2n = 18 – 3n

n = 7

Hence, the 7th term of the both the A.Ps are equal.

Question 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: 

In order the find the 12th term from the end of an A.P which has n terms, it is done by simply finding the ((n -12) + 1)th of the A.P.

As we know that, to find nth term in an A.P = a + (n – 1)d

(i) Given that,

A.P = 3, 5, 7, 9, …. 201



last term is 201

where, a = 3 and d = (5 – 3) = 2

an = 3 + (n – 1)2 = 201

3 + 2n – 2 = 201

2n = 200

n = 100

Hence the A.P has 100 terms.

therefore, the 12th term from the end is same as (100 – 12 + 1)th of the A.P which is the 89th term.

a89 = 3 + (89 – 1)2

= 3 + 88(2)

= 3 + 176 = 179

Hence, the 12th term from the end of the A.P is 179.

(ii) Given that,

A.P = 3,8,13, … ,253

last term is 253

where, a = 3 and d = (8 – 3) = 5

an = 3 + (n – 1)5 = 253

3 + 5n – 5 = 253

5n = 253 + 2 = 255

n = 255/5

n = 51

Hence, the A.P has 51 terms.

therefore, the 12th term from the end is same as (51 – 12 + 1)th of the A.P which is the 40th term.

a40 = 3 + (40 – 1)5

= 3 + 39(5)

= 3 + 195 = 198

Hence, the 12th term from the end of the A.P is 198.

(iii) Given that,

A.P = 1, 4, 7, 10, … ,88

where, a = 1 and d = (4 – 1) = 3

last term is 88

an = 1 + (n – 1)3 = 88

1 + 3n – 3 = 8

3n = 90

n = 30

Hence, the A.P has 30 terms.

therefore, the 12th term from the end is same as (30 – 12 + 1)th of the A.P which is the 19th term.

= a89 = 1 + (19 – 1)3

= 1 + 18(3) = 1 + 54 = 55

Hence the 12th term from the end of the A.P is 55.




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