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 

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, x_{1}, x_{2}, x_{3}, 28 are in AP then x_{3} = ?
(a) 19
(b) 23
(c) 22
(d) cannot be determined
Solution
(c) 22
Here, a = 4, l = 28 and n = 5
Then, T_{5} = 28
⇒ a + (n – 1)d = 28
⇒ 4 + (5 – 1)d = 28
⇒ 4d = 24
⇒ d = 6
Hence, x_{3} = 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, a_{n} = 2n + 1 (Given)
∴ First term, a_{1} = 2 × 1 + 1 = 2 + 1 = 3
Second term, a_{2} = 2 × 2 + 1 = 4 + 1 = 5
Third term, a_{3} = 2 × 3 + 1 = 6 + 1 = 7
∴ Sum of the first three terms a_{1} + a_{2} + a_{3} = 3 + 5 + 7 = 15
6. The sum of first n terms of an AP is (3n^{2} + 6n). The common difference of the AP is
(a) 6
(b) 9
(c) 15
(d) 3
Solution
Let S_{n} denotes the sum of first n terms of the AP.
∴ S_{n} = 3n^{2 }+ 6n
⇒ S_{n1} = 3(n 1)^{2} + 6(n – 1)
= 3(n^{2} – 2n + 1) + 6(n – 1)
= 3n^{2} – 3
So,
n^{th} term of the AP, a_{n} = S_{n} – S_{n1}
= (3n^{2 }+ 6n) – (3n^{2 }– 3)
= 6n + 3
Let d be the common difference of the AP.
∴ d = a_{n} – a_{n1}
= (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 – n^{2}). The nth term of the AP is
(a) (5 – 2n)
(b) (6 – 2n)
(c) (2n – 5)
(d) (2n – 6)
Solution
Let S_{n} denotes the sum of first n terms of the AP.
∴ S_{n }= 5n – n^{2}
⇒ S_{n1} = 5(n – 1) – (n – 1)^{2}
= 5n – 5 – n^{2} + 2n – 1
= 7n – n^{2} – 6
∴ n^{th} term of the AP, a_{n} = S_{n }– S_{n1}
= (5n – n^{2}) – (7n – n^{2} – 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 (4n^{2} + 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 S_{n} denotes the sum of first n terms of the AP.
∴ S_{n} = 4n^{2} + 2n
⇒ S_{n1} = 4(n – 1)^{2} + 2(n – 1)
= 4(n^{2} – 2n + 1) + 2(n – 1)
= 4n^{2} – 6n + 2.
∴ n^{th} term of the AP, a_{n} = S_{n} – S_{n1}
= (4n^{2} + 2n) – (4n^{2} – 6n + 2)
= 8n – 2
Thus, the n^{th} term of the AP is (8n – 2)
9. The 7^{th}term of an AP is 1 and its 16^{th}term 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, a_{n} = a + (n – 1)d
Now,
a_{7} = 1 (Given)
⇒ a + 6d =  1 ...(1)
Also,
a_{16} = 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, a_{n} = 13 + (n – 1) × 2
= 2n – 15
10. The 5^{th}term 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
a_{5} = 3 (Given)
⇒ a + (5 – 1) × (4) = 3 [a_{n} = a + (n – 1)d]
⇒ a – 16 = 3
⇒ a = 16 – 3 = 13
Using the formula, S_{n }= n/2[2a + (n – 1)d], we get
S_{10} = 10/2[2 × 13 + (10 – 1) × (4)]
= 5 × (26 – 36)
= 5 × (10)
=  50
Thus, the sum of its 10 terms is –50.
11. The 5^{th}term of an AP is 20 and the sum of its 7^{th}and 11^{th} 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,
a_{5} = 20
⇒ a + (5 – 1)d = 20 [a_{n} = a + (n – 1)d]
⇒ a + 4d = 20 ...(1)
Now,
a_{7} + a_{11} = 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 13^{th} term of an AP is 4 times its 3^{rd} term. If its 5^{th} 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,
a_{13} = 4 × a_{13} (Given)
⇒ a + 12d = 4a + 8d
⇒ 3a = 4d ....(1)
Also,
a_{5} = 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, S_{n} = n/2[2a + (n – 1)d], we get
S_{10} = 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 a_{50}.
a_{50} = 5 + (50 – 1) × 7 [a_{n }= 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) [S_{n} = 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, S_{n} = n/2[2a + (n – 1)d], we get
S_{40} = 40/2[2 × 6 + (40 – 1) × 6
= 20(12 + 234)
= 20 × 246
= 4920
Thus, the required sum is 4920.
6. How many twodigit numbers are divisible by 3?
(a) 25
(b) 30
(c) 32
(d) 36
Solution
(b) 30
The twodigit 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,
a_{n} = 99
⇒ 12 + (n – 1) × 3 = 99 [a_{n} = a + (n – 1)d]
⇒ 3n + 9 = 99
⇒ 3n = 99 – 9 = 90
⇒ n = 30
Thus, there are 30 twodigit numbers divisible by 3.
17. How many threedigit number are divisible by 9?
(a) 86
(b) 90
(c) 96
(d) 100
Solution
(d) 100
The threedigit 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,
a_{n} = 999
⇒ 108 + (n – 1) × 9 = 999 [a_{n} = a + (n – 1)d]
⇒ 9n + 99 = 999
⇒ 9n = 999 – 99 = 900
⇒ n = 100
Thus, there are 100 threedigit numbers divisible by 9.
18. What is the common difference of an AP in which a_{18} – a_{14} = 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,
a_{18} – a_{14} = 32
⇒ (a + 17d) – (a + 13d) = 32 [a_{n} = a + (n – 1)d]
⇒ 4d = 32
⇒ d = 8
Thus, the common difference of the AP is 8.
19. If a_{n} denotes the nth term of AP 3, 8, 13, 18, .... then what is the value of (a_{30} – a_{20}) ?
(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
∴ a_{30} – a_{20}
= [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) 8^{th}
(b) 9^{th}
(c) 10^{th}
(d) 11^{th}
Solution
(b) 9^{th}
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_{n }= 0
⇒ 72 + (n – 1) × (9) = 0 [a_{n} = a + (n – 1)d]
⇒ 9n + 81 = 0
⇒ n = 81/9 = 9
Thus, the 9^{th} term of the given AP is 0.
21. Which term of the AP 25, 20, 15, .... is the first negative term?
(a) 10^{th}
(b) 9^{th}
(c) 8^{th}
(d) 7^{th}
Solution
(d) 7^{th}
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,
a_{n} < 0
⇒ 25 + (n – 1) × (5) <0 [a_{n} = a + (n – 1)d]
⇒ 30 – 5n < 0
⇒  5n <  30
⇒ n > 30/5 = 6
∴ n = 7
Thus, the 7^{th} term is the first negative term of the given AP.
22. Which term of the AP 21, 42, 63, 84, .... is 210?
(a) 9^{th}
(b) 10^{th}
(c) 11^{th}
(d) 12^{th}
Solution
(b) 10^{th}
Here, a = 21 and d = (42 – 21) = 21
Let 210 be the nth term of the given AP.
Then, T_{n} = 210
⇒ a + (n – 1)d = 210
⇒ 21 + (n – 1) × 21 = 210
⇒ 21n = 210
⇒ n = 10
Hence, 210 is the 10^{th} term of the AP.
23. What is 20^{th} 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 rewrite the given AP in reverse order i.e., 253, 248, ...13, 8, 3.,
Now, the 20^{th} term from the end of the given AP is equal to the 20^{th} term from beginning of the AP 253, 248, ..., 13, 8, 3.
Here, a = 253 and d = 248 – 253 =  5
∴ 20^{th} term of this AP
= 253 + (20 – 1) × (5)
= 253 – 95
= 158
Thus, the 20^{th} 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_{n }= 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 }= n/2[2a + (n – 1)d], we get
S_{16} = 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:
S_{n} = n/2[2a + (n – 1)d] = 406
⇒ n/2[2 × 3 + (n – 1) × 4] = 406
⇒ n[3 + 2n – 2] = 406
⇒ 2n^{2} – 28n + 29n – 406
⇒ 2n2 + n – 406 = 0
⇒ 2n^{2} – 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 2^{nd} term of an AP is 13 and 5^{th} term is 25. What is its 17 term?
(a) 69
(b) 73
(c) 77
(d) 81
Solution
(b) 73
T_{2} = a + d = 13 ...(i)
T_{5} = 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, T_{17} = a + 16d = 9 + 16 × 4 = 73
Hence, the 17^{th} term is 73.
28. The 17^{th} term of an AP exceeds its 10^{th} term by 21. The common difference of the AP is
(a) 3
(b) 2
(c) 3
(d) 2
Solution
(a) 3
T_{10} = a + 9d
T_{17 }= a + 16d
Also, a + 16d = 21 + T_{10}
⇒ a + 16d = 21 + 9d
⇒ 7d = 21
⇒ d = 3
Hence, the common difference of the AP is 3.
29. The 8^{th} term of an AP is 17 and its 14^{th} term is 29. The common difference of the AP is
(a) 3
(b) 2
(c) 5
(d) 2
Solution
(b) 2
T_{8} = a + 7d = 17 ...(i)
T_{14 }= a + 13d = 29 ...(ii)
On subtracting (i) from (ii), we get:
⇒ 6d = 12
⇒ d = 12
Hence, the common difference is 2.
30. The 7^{th} 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
T_{7} = a + 6d
⇒ a + 6 × (4) = 4
⇒ a = 4 + 24 = 28
Hence, the first term us 28.