# RS Aggarwal Solutions Chapter 11 Arithmetic Progression MCQ Class 10 Maths Chapter Name RS Aggarwal Chapter 11 Arithmetic Progression Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 11AExercise 11BExercise 11CExercise 11D Related Study NCERT Solutions for Class 10 Maths

### Arithmetic Progressions MCQ Solutions

1. The common difference of the AP 1/p, (1- p)/p, (1 – 2p)/2, ..... is

(a) p

(b) -p

(c) -1

(d) 1

Solution

(c) -1

The given AP is 1/p, (1 – p)/p, (1 – 2p)/2, ....

∴ Common difference, d = (1 – p)/p – 1/p

= (1 – p – 1)/p

= -p/p

= - 1

2. The common difference of the AP 1/3, (1 – 3b)/3, (1 – 6b)/3, .... is

(a) 1/3

(b) -1/3

(c) b

(d) -b

Solution

The given AP is 1/3, (1 – 3b)/p, (1 – 6b)/3, .....

∴ Common difference, d = (1 – 3b)/3 – 1/3

= (1 – 3b - 1

The next term of the AP.

3. The next term of the AP √7, √28, √63, ... is 4. If 4, x1, x2, x3, 28 are in AP then x3 = ?

(a) 19

(b) 23

(c) 22

(d) cannot be determined

Solution

(c) 22

Here, a = 4, l = 28 and n = 5

Then, T5 = 28

⇒ a + (n – 1)d = 28

⇒ 4 + (5 – 1)d = 28

⇒ 4d = 24

⇒ d = 6

Hence, x3 = 28 – 6

= 22

5. If the nth term of an AP is (2n + 1) then the sum of its first three terms is

(a) 6n + 3

(b) 15

(c) 12

(d) 21

Solution

(b) 15

nth term of the AP, an = 2n + 1 (Given)

∴ First term, a1 = 2 × 1 + 1 = 2 + 1 = 3

Second term, a2 = 2 × 2 + 1 = 4 + 1 = 5

Third term, a3 = 2 × 3 + 1 = 6 + 1 = 7

∴ Sum of the first three terms a1 + a2 + a3 = 3 + 5 + 7 = 15

6. The sum of first n terms of an AP is (3n2 + 6n). The common difference of the AP is

(a) 6

(b) 9

(c) 15

(d) -3

Solution

Let Sn denotes the sum of first n terms of the AP.

∴ Sn = 3n+ 6n

⇒ Sn-1 = 3(n -1)2 + 6(n – 1)

= 3(n2 – 2n + 1) + 6(n – 1)

= 3n2 – 3

So,

nth term of the AP, an = Sn – Sn-1

= (3n+ 6n) – (3n– 3)

= 6n + 3

Let d be the common difference of the AP.

∴ d = an – an-1

= (6n + 3) – [6(n – 1) + 3]

= 6n + 3 – 6(n – 1) – 3

= 6

Thus, the common difference of the AP is 6.

7. The sum of first n terms of an AP is (5n – n2). The nth term of the AP is

(a) (5 – 2n)

(b) (6 – 2n)

(c) (2n – 5)

(d) (2n – 6)

Solution

Let Sn denotes the sum of first n terms of the AP.

∴ S= 5n – n2

⇒ Sn-1 = 5(n – 1) – (n – 1)2

= 5n – 5 – n2 + 2n – 1

= 7n – n2 – 6

∴ nth term of the AP, an = S– Sn-1

= (5n – n2) – (7n – n2 – 6)

= 6 – 2n

Thus, the nth term of the AP is (6 – 2n)

8. The sum of the first n terms of an AP is (4n2 + 2n). The nth term of this AP is

(a) (6n – 2)

(b) (7n – 3)

(c) (8n – 2)

(d) (8n + 2)

Solution

(c) (8n – 2)

Let Sn denotes the sum of first n terms of the AP.

∴ Sn = 4n2 + 2n

⇒ Sn-1 = 4(n – 1)2 + 2(n – 1)

= 4(n2 – 2n + 1) + 2(n – 1)

= 4n2 – 6n + 2.

∴ nth term of the AP, an = Sn – Sn-1

= (4n2 + 2n) – (4n2 – 6n + 2)

= 8n – 2

Thus, the nth term of the AP is (8n – 2)

9. The 7thterm of an AP is -1 and its 16thterm is 17. The nth term of the AP is

(a) (3n + 8)

(b) (4n – 7)

(c) (15 – 2n)

(d) (2n – 15)

Solution

Let a be the first term and d be the common difference of the AP. Then, nth term of the AP, an = a + (n – 1)d

Now,

a7 = -1 (Given)

⇒ a + 6d = - 1 ...(1)

Also,

a16 = 17 (Given)

⇒ a + 15d = 17 ...(2)

Subtracting (1) from (2), we get

(a + 15d) – (a + 6d) = 17 – (-1)

⇒ 9d = 18

⇒ d = 2

Putting d = 2 in (1), we get

a + 6 × 2 = - 1

⇒ a = -1 - 12 = - 13

∴ nth term of the AP, an = -13 + (n – 1) × 2

= 2n – 15

10. The 5thterm of an AP is -3 and its common difference is -4. The sum of the first 10 terms is

(a) 50

(b) -50

(c) 30

(d) -30

Solution

(b) -50

Let a be the first term of the AP.

Here, d = - 4

a5 = -3 (Given)

⇒ a + (5 – 1) × (-4) = -3 [an = a + (n – 1)d]

⇒ a – 16 = -3

⇒ a = 16 – 3 = 13

Using the formula, S= n/2[2a + (n – 1)d], we get

S10 = 10/2[2 × 13 + (10 – 1) × (-4)]

= 5 × (26 – 36)

= 5 × (-10)

= - 50

Thus, the sum of its 10 terms is –50.

11. The 5thterm of an AP is 20 and the sum of its 7thand 11th terms is 64. The common difference of the AP is

(a) 4

(b) 5

(c) 3

(d) 2

Solution

(c) 3

Let a be the first term and d be the common difference of the AP. then,

a5 = 20

⇒ a + (5 – 1)d = 20 [an = a + (n – 1)d]

⇒ a + 4d = 20 ...(1)

Now,

a7 + a11 = 64 (Given)

⇒ (a + 6d) + (a + 10d) = 64

⇒ 2a + 16d = 64

⇒ a + 8d = 32 ...(2)

From (1) and (2), we get

20 – 4d + 8d = 32

⇒ 4d = 32 – 20 = 12

⇒ d = 3

Thus, the common difference of the AP is 3.

12. The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16 then the sum of its first ten terms is

(a) 150

(b) 175

(c) 160

(d) 135

Solution

(b) 175

Let a be the first term and d be common difference of the AP. Then,

a13 = 4 × a13 (Given)

⇒ a + 12d = 4a + 8d

⇒ 3a = 4d ....(1)

Also,

a5 = 16 (Given)

⇒ a + 4d = 16 ...(2)

Solving (1) and (2), we get

a + 3a = 16

⇒ 4a = 16

⇒ a = 4

Putting a = 4 in (1), we get

4d = 3 × 4 = 12

⇒ d = 3

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S10 = 10/2[2 × 4 + (10 – 1) × 3]

= 5 × (8 + 27)

= 5 × 35

= 175

Thus, the sum of its first 10 terms is 175.

13. An AP 5, 12, 9, ... has 50 terms. Its last term is

(a) 343

(b) 353

(c) 348

(d) 362

Solution

(c) 348

The given AP is 5, 12, 19, ......

Here, a = 5, d = 12 – 5 = 7 and n = 50

Since there are 50 terms in the AP, so the last term of the AP is a50.

a50 = 5 + (50 – 1) × 7 [a= a + (n – 1)d]

= 5 + 343

= 348

Thus, the last term of the AP is 348.

14. The sum of the first 20 odd natural numbers is

(a) 100

(b) 210

(c) 400

(d) 420

Solution

(c) 400

The first 20 odd natural numbers are 1, 3, 5, ...., 39.

These numbers are in AP.

Here, a = 1, l = 39 and n = 20

∴ Sum of first 20 odd natural numbers

= 20/2(1 + 39) [Sn = n/2(a + l)]

= 10 × 40

= 400

15. The sum of first 40 positive integers divisible by 6 is

(a) 2460

(b) 3640

(c) 4920

(d) 4860

Solution

The positive integers divisible by 6 are 6, 12, 18, .....

This is an AP with a = 6 and d = 6.

Also, n = 40 (Given)

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S40 = 40/2[2 × 6 + (40 – 1) × 6

= 20(12 + 234)

= 20 × 246

= 4920

Thus, the required sum is 4920.

6. How many two-digit numbers are divisible by 3?

(a) 25

(b) 30

(c) 32

(d) 36

Solution

(b) 30

The two-digit numbers divisible by 3 are 12, 15, 18, .....99.

Clearly, these numbers are in AP.

Here, a = 12 and d = 15 – 12 = 3

Let this AP contains n terms. Then,

an = 99

⇒ 12 + (n – 1) × 3 = 99 [an = a + (n – 1)d]

⇒ 3n + 9 = 99

⇒ 3n = 99 – 9 = 90

⇒ n = 30

Thus, there are 30 two-digit numbers divisible by 3.

17. How many three-digit number are divisible by 9?

(a) 86

(b) 90

(c) 96

(d) 100

Solution

(d) 100

The three-digit numbers divisible by 9 are 108, 117, 126, ...... 999.

Clearly, these numbers are in AP.

Here, a = 108 and d = 117 – 108 = 9

Let this AP contains n terms. Then,

an = 999

⇒ 108 + (n – 1) × 9 = 999 [an = a + (n – 1)d]

⇒ 9n + 99 = 999

⇒ 9n = 999 – 99 = 900

⇒ n = 100

Thus, there are 100 three-digit numbers divisible by 9.

18. What is the common difference of an AP in which a18 – a14 = 32?

(a) 8

(b) -8

(c) 4

(d) – 4

Solution

(a) 8

Let a be the first term and d be the common difference of the AP. Then,

a18 – a14 = 32

⇒ (a + 17d) – (a + 13d) = 32 [an = a + (n – 1)d]

⇒ 4d = 32

⇒ d = 8

Thus, the common difference of the AP is 8.

19If an denotes the nth term of AP 3, 8, 13, 18, .... then what is the value of (a30 – a20) ?

(a) 40

(b) 36

(c) 50

(d) 56

Solution

(c) 50

The given AP is 3, 8, 13, 18.....

Here, a = 3 and d = 8 – 3 = 5

∴ a30 – a20

= [3 + (30 – 1) × 5] – [3 + (20 – 1) × 5] [an = a + (n – 1)d]

= 148 – 98

= 50

Thus, the required value is 50.

20. Which term of the AP 72, 63, 54, .... is 0?

(a) 8th

(b) 9th

(c) 10th

(d) 11th

Solution

(b) 9th

The given AP is 72, 63, 54, .....

Here, a = 72 and d = 63 – 72 = -9

Suppose nth term of the given AP is 0. Then,

a= 0

⇒ 72 + (n – 1) × (-9) = 0 [an = a + (n – 1)d]

⇒ -9n + 81 = 0

⇒ n = 81/9 = 9

Thus, the 9th term of the given AP is 0.

21. Which term of the AP 25, 20, 15, .... is the first negative term?

(a) 10th

(b) 9th

(c) 8th

(d) 7th

Solution

(d) 7th

The given AP is 25, 20, 15, .....

Here, a = 25 and d = 20 – 25 = - 5

Let the nth term of the given AP be the first negative term. Then,

an < 0

⇒ 25 + (n – 1) × (-5) <0  [an = a + (n – 1)d]

⇒ 30 – 5n < 0

⇒ - 5n < - 30

⇒ n > 30/5 = 6

∴ n = 7

Thus, the 7th term is the first negative term of the given AP.

22. Which term of the AP 21, 42, 63, 84, .... is 210?

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Solution

(b) 10th

Here, a = 21 and d = (42 – 21) = 21

Let 210 be the nth term of the given AP.

Then, Tn = 210

⇒ a + (n – 1)d = 210

⇒ 21 + (n – 1) × 21 = 210

⇒ 21n = 210

⇒ n = 10

Hence, 210 is the 10th term of the AP.

23. What is 20th term from the end of the AP 3, 8, 13, ....., 253?

(a) 163

(b) 158

(c) 153

(d) 148

Solution

(b) 158

The given AP is 3, 8, 13, 253.

Let us re-write the given AP in reverse order i.e., 253, 248, ...13, 8, 3.,

Now, the 20th term from the end of the given AP is equal to the 20th term from beginning of the AP 253, 248, ..., 13, 8, 3.

Here, a = 253 and d = 248 – 253 = - 5

∴ 20th term of this AP

= 253 + (20 – 1) × (-5)

= 253 – 95

= 158

Thus, the 20th term from the end of the given AP is 158.

24. (5 + 13 + 21 + ......+ 181) = ?

(a) 2476

(b) 2337

(c) 2219

(d) 2139

Solution

(d) 2139

Here, a = 5, d = (13 – 5) = 8 and l = 181

Let the number of terms be n.

Then, T= 181

⇒ a + (n – 1)d = 181

⇒ 5 + (n – 1) × 8 = 181

⇒ 8n = 184

⇒ n = 23

∴ Required sum = n/2(a + l)

= 23/2(5 + 181)

= 23 × 93

= 2139.

Hence, the required sum is 2139.

25. The sum of first 16 terms of the AP 10, 6, 2, ... is

(a) 320

(b) -320

(c) -352

(d) -400

Solution

(b) -320

Here, a = 10, d = (6 – 10) = - 4 and n = 16

Using the formula, S= n/2[2a + (n – 1)d], we get

S16 = 16/2[2 × 10 + (16 – 1) × (-4)]

[∵ a = 10, d = - 4 and n = 16]

= 8 × [20 – 60]

= 8 × (-40)

= - 320

Hence, the sum of the first 16 terms of the given AP is -320.

26. How many terms of the AP 3, 7, 11, 15, … will make the sum 406 ?

(a) 10

(b) 12

(c) 14

(d) 20

Solution

(c) 14

Here, a = 3 and d = (7 – 3) = 4

Let the sum of n terms be 406.

Then, we have:

Sn = n/2[2a + (n – 1)d] = 406

⇒ n/2[2 × 3 + (n – 1) × 4] = 406

⇒ n[3 + 2n – 2] = 406

⇒ 2n2 – 28n + 29n – 406

⇒ 2n2 + n – 406 = 0

⇒ 2n2 – 28n + 29n – 406 = 0

⇒ 2n(n – 14) + 29(n – 14) = 0

⇒ (2n + 29)(n – 14) = 0

⇒ n = 14 (∵ n can’t be a fraction)

Hence, 14 terms will make the sum 406.

27. The 2nd term of an AP is 13 and 5th term is 25. What is its 17 term?

(a) 69

(b) 73

(c) 77

(d) 81

Solution

(b) 73

T2 = a + d = 13 ...(i)

T5 = a + 4d = 25 ...(ii)

On subtracting (i) from (ii), we get:

⇒ 3d = 12

⇒ d = 4

On putting the value of d in (i), we get:

⇒ 3d = 12

⇒ d = 4

On putting the value of d in (i), we get:

⇒ a + 4 = 13

⇒ a = 9

Now, T17 = a + 16d = 9 + 16 × 4 = 73

Hence, the 17th term is 73.

28. The 17th term of an AP exceeds its 10th term by 21. The common difference of the AP is

(a) 3

(b) 2

(c) -3

(d) -2

Solution

(a) 3

T10 = a + 9d

T17 = a + 16d

Also, a + 16d = 21 + T10

⇒ a + 16d = 21 + 9d

⇒ 7d = 21

⇒ d = 3

Hence, the common difference of the AP is 3.

29. The 8th term of an AP is 17 and its 14th term is 29. The common difference of the AP is

(a) 3

(b) 2

(c) 5

(d) -2

Solution

(b) 2

T8 = a + 7d = 17 ...(i)

T14 = a + 13d = 29 ...(ii)

On subtracting (i) from (ii), we get:

⇒ 6d = 12

⇒ d = 12

Hence, the common difference is 2.

30. The 7th term of an AP is 4 and its common difference is -4. What is its first term?

(a) 16

(b) 20

(c) 24

(d) 28

Solution

(d) 28

T7 = a + 6d

⇒ a + 6 × (-4) = 4

⇒ a = 4 + 24 = 28

Hence, the first term us 28.