Arithmetic Progressions: 1 mark Questions (CBSE Class 10)
77 different 1 mark questions on Arithmetic Progressions from CBSE Class 10 Maths board exams 2022–2026, newest first.
The sum of first n terms of an A.P. is 50 2 . If the first and the last terms are 2 and 19 2 respectively, then the value of n is :
(A) 10(B) 5(C) 15(D) 20
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Answer: (B) 5
S n = 2 n ( a + l ) .50 2 = 2 n ( 2 + 19 2 ) = 10 n 2 .n = 5 .
n t h term of the A.P. − 2 3 , 2 3 , 2 9 , ... is :
(A) 2 3 n − 3 (B) 3 n − 2 9 (C) 2 3 n − 9 (D) 3 n + 2 3
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Answer: (B) 3 n − 2 9
a = − 2 3 , d = 2 3 − ( − 2 3 ) = 3 .a n = − 2 3 + ( n − 1 ) 3 = 3 n − 2 9 .
n t h term of an A.P. is 5n – 15. The common difference of the A.P. is :
(A) 5n(B) 5(C) – 5(D) 10
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Answer: (B) 5
d = a n + 1 − a n = [ 5 ( n + 1 ) − 15 ] − [ 5 n − 15 ] = 5 .
n t h term of the A.P. : 3 − 1 , 3 4 , 3 , … is
(A) 3 5 n − 9 (B) 3 5 n − 6 (C) 3 3 n − 4 (D) 3 3 n + 2
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Answer: (B) 3 5 n − 6
a = − 3 1 , d = 3 4 − ( − 3 1 ) = 3 5 a n = − 3 1 + ( n − 1 ) 3 5 = 3 5 n − 6
If − 26 , x , 2 are in A.P., then the value of x is
(A) 14(B) − 13 (C) − 12 (D) − 14
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Answer: (C) − 12
For an A.P., 2 x = − 26 + 2 = − 24 . x = − 12
If 9 − 20 , 9 − 2 , 9 16 , … are in A.P., then next term of the sequence is
(A) 9 32 (B) 9 46 (C) 9 2 (D) 9 34
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Answer: (D) 9 34
d = 9 − 2 − 9 − 20 = 9 18 = 2 Next term = 9 16 + 9 18 = 9 34
If 1 4 t h term of an A.P. is 4 and its 1 5 t h term is zero, then its first term is
(A) − 48 (B) − 56 (C) 56(D) 48
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Answer: (C) 56
d = a 15 − a 14 = 0 − 4 = − 4 a 14 = a + 13 d ⇒ 4 = a − 52 ⇒ a = 56
The common difference of the AP : 2 , 2 2 , 3 2 , 4 2 , ..... is :
(A) 2 (B) 1(C) 2 2 (D) − 2
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d = a 2 − a 1 = 2 2 − 2 = 2 .
The first term of an AP is p and the common difference is q , then its 10th term is :
(A) q − 9 p (B) p − 9 q (C) p + 9 q (D) 2 p + 9 q
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Answer: (C) p + 9 q
a n = a + ( n − 1 ) d .a 10 = p + 9 q .
Which of the following sequence is not an A.P. ?
(A) 2 , 2 5 , 3 , 2 7 , ... (B) − 1.2 , − 3.2 , − 5.2 , − 7.2 , ... (C) 2 , 8 , 18 , ... (D) 1 2 , 3 2 , 5 2 , 7 2 , ...
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Answer: (D) 1 2 , 3 2 , 5 2 , 7 2 , ...
(A) 2 , 2 5 , 3 , 2 7 : common difference 2 1 , an A.P. (B) − 1.2 , − 3.2 , − 5.2 , − 7.2 : common difference − 2 , an A.P. (C) 2 , 2 2 , 3 2 : common difference 2 , an A.P. (D) 1 , 9 , 25 , 49 : differences 8, 16, 24 are not equal, so it is not an A.P.
Assertion (A) : The mean of first 'n' natural numbers is 2 n − 1 . Reason (R): The sum of first 'n' natural numbers is 2 n ( n + 1 ) .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
Sum of first n natural numbers = 2 n ( n + 1 ) , so R is true. Mean = 2 n ( n + 1 ) ÷ n = 2 n + 1 , not 2 n − 1 , so A is false.
The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P. is :
(A) 6(B) − 6 (C) 18(D) − 18
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Answer: (A) 6
For three consecutive terms of an A.P., 2 ( x + 10 ) = 2 x + ( 3 x + 2 ) . 2 x + 20 = 5 x + 2 3 x = 18 , so x = 6
The number of multiples of 4 lying between 12 and 250 is :
(A) 59(B) 59.5(C) 60(D) 61
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Answer: (A) 59
Multiples of 4 between 12 and 250: 16, 20, ..., 248 248 = 16 + ( n − 1 ) × 4 , so n − 1 = 58 , n = 59
The n t h term of the A.P. 3 − 1 , 3 2 , 3 5 , 3 8 , ... is :
(A) 3 n − 4 (B) n − 3 4 (C) 3 n − 2 (D) 3 n − 4
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Answer: (B) n − 3 4
a = − 3 1 , d = 3 2 − ( − 3 1 ) = 1 a n = − 3 1 + ( n − 1 ) × 1 = n − 3 4
The number of multiples of 6 lying between 25 and 363 is :
(A) 56(B) 56.5(C) 57(D) 58
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Answer: (A) 56
Multiples of 6 between 25 and 363: 30, 36, ..., 360 360 = 30 + ( n − 1 ) × 6 , so n − 1 = 55 , n = 56
The n t h term of an A.P. is 2 n + 1 . Its common difference is
(A) 2 (B) 2 n (C) 1(D) 2 + 1
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d = a n + 1 − a n = 2 ( n + 1 ) + 1 − ( 2 n + 1 ) = 2
In an A.P., a = − 3 and S 17 = 357 . The value of a 17 is
(A) 47(B) 39(C) 45(D) 42
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Answer: (C) 45
S 17 = 2 17 ( a + a 17 ) = 357 a + a 17 = 42 ⇒ a 17 = 42 + 3 = 45
If sum of first ten terms of an A.P. is zero with a as the first term and d, the common difference, which of the following relation is true ?
(A) 10 a + 9 d = 0 (B) 2 a = 9 d (C) a 10 = − a (D) a 10 = a
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Answer: (C) a 10 = − a
S 10 = 2 10 ( 2 a + 9 d ) = 0 ⇒ 2 a + 9 d = 0 ⇒ 9 d = − 2 a a 10 = a + 9 d = a − 2 a = − a
If a n represents n t h term of the A.P. − 4 15 , − 4 10 , − 4 5 , … then value of a 16 − a 12 is
(A) 4(B) 4 5 (C) 5(D) 4 25
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Answer: (C) 5
d = − 4 10 − ( − 4 15 ) = 4 5 a 16 − a 12 = ( a + 15 d ) − ( a + 11 d ) = 4 d = 4 × 4 5 = 5
In an A.P., if a 14 − a 8 = 24 , then the common difference of the A.P. is
(A) 6(B) 4(C) ± 4 (D) 3
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Answer: (B) 4
a 14 − a 8 = ( a + 13 d ) − ( a + 7 d ) = 6 d = 24 d = 4
If n t h term of an A.P. is 5 n − 6 , then its common difference is :
(A) − 6 (B) 5 n (C) 5(D) 6
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Answer: (C) 5
a n = 5 n − 6 , so a n + 1 = 5 ( n + 1 ) − 6 = 5 n − 1 .d = a n + 1 − a n = ( 5 n − 1 ) − ( 5 n − 6 ) = 5 .
The 2 0 t h term of the A.P. : 10 2 , 6 2 , 2 2 , .... is :
(A) − 76 + 10 2 (B) − 62 2 (C) − 66 2 (D) 86 2
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a = 10 2 , d = 6 2 − 10 2 = − 4 2 .a 20 = a + 19 d = 10 2 − 76 2 = − 66 2 .
In an A.P., a n − a n − 4 = 32 . Its common difference is :
(A) − 8 (B) 8(C) 4 n (D) 4
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Answer: (B) 8
a n − a n − 4 = [ a + ( n − 1 ) d ] − [ a + ( n − 5 ) d ] = 4 d .4 d = 32 , so d = 8 .
1 0 t h term of the A.P. : –12, –19, –26, .... is
(A) –75(B) –65(C) 51(D) –82
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Answer: (A) –75
a = − 12 , d = − 19 − ( − 12 ) = − 7 .a 10 = a + 9 d = − 12 + 9 ( − 7 ) = − 12 − 63 = − 75 .
The 1 6 t h term of the A.P. : 5 3 , 2 3 , − 3 , ... is
(A) − 25 3 (B) − 40 3 (C) 50 3 (D) − 45 + 5 3
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a = 5 3 , d = 2 3 − 5 3 = − 3 3 .a 16 = a + 15 d = 5 3 − 45 3 = − 40 3 .
2 2 n d term of the A.P. : 2 3 , 2 1 , 2 − 1 , 2 − 3 , ........ is
(A) 2 45 (B) − 9 (C) 2 − 39 (D) − 21
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Answer: (C) 2 − 39
a = 2 3 , d = 2 1 − 2 3 = − 1 .a 22 = a + 21 d = 2 3 − 21 = 2 − 39 .
If the sum of first n terms of an A.P. is given by S n = 2 n ( 3 n + 1 ) , then the first term of the A.P. is
(A) 2(B) 2 3 (C) 4(D) 2 5
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Answer: (A) 2
First term a = S 1 = 2 1 ( 3 + 1 ) = 2 .
1 5 t h term of the A.P. 3 13 , 3 9 , 3 5 , ....... is
(A) 23(B) 3 − 53 (C) − 11 (D) 3 − 43
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Answer: (D) 3 − 43
a = 3 13 , d = 3 9 − 3 13 = − 3 4 .a 15 = a + 14 d = 3 13 − 3 56 = 3 − 43 .
If the sum of first m terms of an AP is 2 m 2 + 3 m , then its second term is :
(A) 10 (B) 9 (C) 12 (D) 4
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Answer: (B) 9
S 1 = 2 + 3 = 5 , so a 1 = 5 .S 2 = 8 + 6 = 14 .a 2 = S 2 − S 1 = 14 − 5 = 9 .
The 1 1 th and 1 3 th term of an AP are 39 and 45 , respectively. What is the common difference of the AP ?
(A) 42 (B) 21 (C) 6 (D) 3
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Answer: (D) 3
a 13 − a 11 = ( a + 12 d ) − ( a + 10 d ) = 2 d .2 d = 45 − 39 = 6 , so d = 3 .
The 1 0 t h term of the AP5 , 4 19 , 2 9 , 4 17 , … is :
(A) 4 11 (B) 11 4 (C) 4 13 (D) 13 4
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Answer: (A) 4 11
a = 5 , d = 4 19 − 5 = − 4 1 a 10 = a + 9 d = 5 − 4 9 = 4 11
Assertion (A) : Common difference of the AP : 5 , 1 , − 3 , − 7 , … is 4. Reason (R) : Common difference of the AP : a 1 , a 2 , a 3 , … , a n is obtained by d = a n − a n − 1 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
Common difference = 1 − 5 = − 4 , not 4, so A is false d = a n − a n − 1 is the correct definition, so R is true
The 9 t h term from the end (towards first term) of the AP 7 , 11 , 15 , 19 , … , 147 is :
(A) 135(B) 125(C) 115(D) 39
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Answer: (C) 115
Reading from the end, the AP is 147 , 143 , 139 , … with first term 147 and d = − 4 9 t h term from the end = 147 + 8 × ( − 4 ) = 147 − 32 = 115
Assertion (A) : For an A.P., 3,6,9, ..., 198, 1 0 th term from the end is 168. Reason (R) : If 'a' and 'l ' are the first term and last term of an A.P. with common difference 'd', then n th term from the end of the given A.P. is l − ( n − 1 ) d .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
R is the standard result: n th term from the end = l − ( n − 1 ) d . R is true. Here l = 198 , d = 3 : 1 0 th term from the end = 198 − 9 × 3 = 171 = 168 . A is false.
In an A.P. ; if a = 8 and a 10 = − 19 , then value of d is :
(A) 3(B) − 9 11 (C) − 10 27 (D) − 3
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Answer: (D) − 3
a 10 = a + 9 d .− 19 = 8 + 9 d , so 9 d = − 27 and d = − 3 .
In an A.P., if d = − 4 and a 7 = 4 , then the first term ‘a’ is equal to
(A) 6(B) 7(C) 20(D) 28
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Answer: (D) 28
a 7 = a + 6 d 4 = a + 6 ( − 4 ) = a − 24 a = 28
If the first term of an AP is − 3 and common difference − 2 , then the seventh term is
(A) − 9 (B) 9(C) − 17 (D) − 15
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Answer: (D) − 15
a 7 = a + 6 d = − 3 + 6 ( − 2 ) = − 3 − 12 = − 15
The common difference of an A.P. whose n t h term is given by a n = 5 n − 1 , is :
(A) 1(B) 6(C) 5(D) 4
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Answer: (C) 5
a 1 = 4 , a 2 = 9 .d = a 2 − a 1 = 5 .
The 3 0 t h term of the A.P. − 3 , − 7 , − 11 , … is :
(A) 113(B) − 117 (C) − 119 (D) − 120
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Answer: (C) − 119
a = − 3 , d = − 7 − ( − 3 ) = − 4 .a 30 = a + 29 d = − 3 + 29 × ( − 4 ) = − 3 − 116 = − 119 .
If in an A.P., a = 2 and S 10 = 335 , then its 1 0 t h term is :
(A) 55(B) 65(C) 68(D) 58
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Answer: (B) 65
S 10 = 2 10 ( a + a 10 ) , so 335 = 5 ( 2 + a 10 ) .2 + a 10 = 67 , so a 10 = 65 .
The common difference of an A.P., if a 23 − a 19 = 32 , is :
(A) 8 (B) − 8 (C) − 4 (D) 4
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Answer: (A) 8
a 23 − a 19 = ( a + 22 d ) − ( a + 18 d ) = 4 d .4 d = 32 ⇒ d = 8 .
Assertion (A) : Common difference of the A.P. 5 , 1 , − 3 , − 7 .... is 4. Reason (R) : Common difference of the A.P. a 1 , a 2 , a 3 ...., a n is obtained by d = a n − a n − 1 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
d = a n − a n − 1 is the correct way to find the common difference, so R is true.For 5, 1, − 3 , − 7 , ...: d = 1 − 5 = − 4 , not 4. So A is false.
In an A.P., if the first term a = 7 , n th term a n = 84 and the sum of first n terms s n = 2 2093 , then n is equal to :
(A) 22(B) 24(C) 23(D) 26
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Answer: (C) 23
s n = 2 n ( a + a n ) = 2 n ( 7 + 84 ) = 2 91 n .2 91 n = 2 2093 , so n = 91 2093 = 23 .
n th term of an A.P. is 7 n + 4 . The common difference is :
(A) 7 n (B) 4(C) 7(D) 1
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Answer: (C) 7
d = a n + 1 − a n = [ 7 ( n + 1 ) + 4 ] − ( 7 n + 4 ) = 7 .
In an A.P., if the first term ( a ) = − 16 and the common difference ( d ) = − 2 , then the sum of first 10 terms is :
(A) − 200 (B) − 70 (C) − 250 (D) 250
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Answer: (C) − 250
S 10 = 2 10 [ 2 ( − 16 ) + 9 ( − 2 )] = 5 ( − 32 − 18 ) = 5 × ( − 50 ) = − 250 .
The common difference of an A.P. in which a 15 − a 11 = 48 , is
(A) 12(B) 16(C) − 12 (D) − 16
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Answer: (A) 12
a 15 − a 11 = ( a + 14 d ) − ( a + 10 d ) = 4 d 4 d = 48 , so d = 12 .
Which term of the A.P. − 29 , − 26 , − 23 , … , 61 is 16 ?
(A) 1 1 t h (B) 1 6 t h (C) 1 0 t h (D) 3 1 s t
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Answer: (B) 1 6 t h
a = − 29 , d = 3 a n = − 29 + ( n − 1 ) 3 = 16 3 ( n − 1 ) = 45 , so n = 16 .
The sum of first 200 natural numbers is
(A) 2010(B) 2000(C) 20100(D) 21000
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Answer: (C) 20100
S n = 2 n ( n + 1 ) S 200 = 2 200 × 201 = 20100
The common difference of an A.P. in which a 20 − a 15 = 20 , is
(A) 4(B) 5(C) 4d(D) 5d
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Answer: (A) 4
a 20 − a 15 = ( a + 19 d ) − ( a + 14 d ) = 5 d 5 d = 20 , so d = 4 .
The common difference of the A.P.2 x 1 , 2 x 1 − 4 x , 2 x 1 − 8 x , ................is :
(A) − 2 x (B) − 2 (C) 2 (D) 2 x
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Answer: (B) − 2
d = 2 x 1 − 4 x − 2 x 1 = 2 x − 4 x = − 2 .Check: 2 x 1 − 8 x − 2 x 1 − 4 x = 2 x − 4 x = − 2 . So d = − 2 .
If the sum of first n terms of an A.P. is 3 n 2 + 4 n and its common difference is 6, then its first term is :
(A) 7(B) 4(C) 6(D) 3
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Answer: (A) 7
First term = S 1 = 3 ( 1 ) 2 + 4 ( 1 ) = 7 . (Check: S 2 = 20 , so a 2 = 13 and d = 13 − 7 = 6 , as given.)
Three numbers in A.P. have the sum 30. What is its middle term ?
(A) 4(B) 10(C) 16(D) 8
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Answer: (B) 10
Let the numbers be a − d , a , a + d . Sum = 3 a = 30 , so a = 10 . Middle term = 10.
If the first three terms of an A.P. are 3 p − 1 , 3 p + 5 , 5 p + 1 respectively; then the value of p is :
(A) 2(B) − 3 (C) 4(D) 5
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Answer: (D) 5
For an A.P., 2 × (middle term) = sum of the other two. 2 ( 3 p + 5 ) = ( 3 p − 1 ) + ( 5 p + 1 ) .6 p + 10 = 8 p , so p = 5 .
The next (4 t h ) term of the A.P. 18 , 50 , 98 , … is :
(A) 128 (B) 140 (C) 162 (D) 200
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18 = 3 2 , 50 = 5 2 , 98 = 7 2 , so d = 2 2 .Next term = 7 2 + 2 2 = 9 2 = 162 .
The number of terms in the A.P. 3, 6, 9, 12, ..., 111 is :
(A) 36(B) 40(C) 37(D) 30
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Answer: (C) 37
a = 3 , d = 3 , a n = 111 .3 + ( n − 1 ) 3 = 111 ⇒ 3 n = 111 ⇒ n = 37 .
The next (4 t h ) term of the A.P. 7 , 28 , 63 , … is :
(A) 70 (B) 84 (C) 97 (D) 112
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7 , 28 = 2 7 , 63 = 3 7 , so d = 7 .Next term = 3 7 + 7 = 4 7 = 112 .
The 1 4 t h term from the end of the A.P. –11, –8, –5, ..., 49 is :
(A) 7(B) 10(C) 13(D) 28
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Answer: (B) 10
Read from the end, the A.P. is 49, 46, 43, ... with first term 49 and common difference –3. 1 4 t h term from the end = 49 + 13 × ( − 3 ) = 49 − 39 = 10 .
The seventh term of an A.P. whose first term is 28 and common difference − 4 , is
(A) 0(B) 4(C) 52(D) 56
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Answer: (B) 4
a 7 = a + 6 d = 28 + 6 ( − 4 ) = 28 − 24 = 4
In an AP, if d = − 4 , n = 7 and a n = 4 , then the value of a is
(A) 6(B) 7(C) 20(D) 28
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Answer: (D) 28
a n = a + ( n − 1 ) d 4 = a + 6 ( − 4 ) = a − 24 a = 28
How many terms are there in the A.P. given below ? 14, 19, 24, 29, ....., 119
(A) 18(B) 14(C) 22(D) 21
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Answer: (c) 22
Here a = 14, d = 5 and the last term is 119. 14 + ( n − 1 ) 5 = 119 ( n − 1 ) 5 = 105 , so n − 1 = 21 n = 22
The sum of the first 21 terms of an A.P. : 16, 12, 8, 4, ..... is :
(A) − 480 (B) − 504 (C) 1176(D) − 484
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Answer: (b) − 504
a = 16, d = −4, n = 21 S 21 = 2 21 [ 2 × 16 + 20 × ( − 4 )] = 2 21 × ( − 48 ) = − 504
The 8 t h term of an A.P. is 17 and its 1 4 t h term is 29. The common difference of this A.P. is :
(A) 3(B) 2(C) 5(D) − 2
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Answer: (B) 2
a + 7 d = 17 and a + 13 d = 29 .Subtracting, 6 d = 12 , so d = 2 .
The sum of the first 100 even natural numbers is :
(A) 10100(B) 2550(C) 5050(D) 10010
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Answer: (A) 10100
The numbers are 2, 4, ..., 200 with n = 100 . S = 2 100 ( 2 + 200 ) = 50 × 202 = 10100 .
The sum of the first 50 odd natural numbers is :
(A) 5000(B) 2500(C) 2550(D) 5050
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Answer: (B) 2500
The numbers are 1, 3, 5, ..., 99 with a = 1 , d = 2 , n = 50 . S 50 = 2 50 ( 1 + 99 ) = 25 × 100 = 2500 .
If − 5 , x , 3 are three consecutive terms of an A.P., then the value of x is
(A) − 2 (B) 2(C) 1(D) − 1
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Answer: (D) − 1
For consecutive terms of an A.P., 2 x = − 5 + 3 = − 2 . So x = − 1 .
The 2 0 th term of an A.P, whose first term is − 2 and the common difference is 4, is
(A) 78(B) 74(C) − 36 (D) − 34
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Answer: (B) 74
a n = a + ( n − 1 ) d .a 20 = − 2 + 19 × 4 = − 2 + 76 = 74 .
If p − 1 , p + 1 and 2 p + 3 are in A.P., then the value of p is
(A) –2(B) 4(C) 0(D) 2
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Answer: (C) 0
For three terms in A.P., 2 ( p + 1 ) = ( p − 1 ) + ( 2 p + 3 ) . 2 p + 2 = 3 p + 2 p = 0
If a, b, c form an A.P. with common difference d, then the value of a − 2 b − c is equal to
(A) 2a + 4d(B) 0(C) –2a – 4d(D) –2a – 3d
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Answer: (C) –2a – 4d
b = a + d and c = a + 2d. a − 2 b − c = a − 2 ( a + d ) − ( a + 2 d ) = a − 2 a − 2 d − a − 2 d = − 2 a − 4 d
The next term of the A.P. : 6 , 24 , 54 is :
(A) 60 (B) 96 (C) 72 (D) 216
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6 , 24 = 2 6 , 54 = 3 6 , so d = 6 .Next term = 4 6 = 16 × 6 = 96
Assertion (A) : a, b, c are in A.P. if and only if 2 b = a + c . Reason (R) : The sum of first n odd natural numbers is n 2 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true but Reason (R) is false.(D) Assertion (A) is false but Reason (R) is true.
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Answer: (B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
a, b, c are in A.P. ⟺ b − a = c − b ⟺ 2 b = a + c . So A is true. 1 + 3 + 5 + ⋯ + ( 2 n − 1 ) = 2 n [ 1 + ( 2 n − 1 )] = n 2 . So R is true.R is not about the condition for three terms in A.P., so it does not explain A.
If k + 2 , 4 k − 6 and 3 k − 2 are three consecutive terms of an A.P., then the value of k is :
(A) 3(B) − 3 (C) 4(D) − 4
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Answer: (A) 3
For three consecutive terms of an A.P., 2 b = a + c . 2 ( 4 k − 6 ) = ( k + 2 ) + ( 3 k − 2 ) 8 k − 12 = 4 k ⇒ k = 3
The next term of the A.P. : 7 , 28 , 63 is :
(A) 70 (B) 80 (C) 97 (D) 112
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7 , 28 = 2 7 , 63 = 3 7 , so d = 7 .Next term = 4 7 = 16 × 7 = 112
The common difference of the A.P. whose n t h term is given by a n = 3 n + 7 , is :
(A) 7(B) 3(C) 3n(D) 1
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Answer: (B) 3
a 1 = 3 ( 1 ) + 7 = 10 , a 2 = 3 ( 2 ) + 7 = 13 .d = a 2 − a 1 = 13 − 10 = 3 .
The 1 1 t h term from the end of the A.P. : 10, 7, 4, ......., − 62 is :
(A) 25(B) 16(C) − 32 (D) 0
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Answer: (C) − 32
Reverse the A.P.: − 62 , − 59 , − 56 , … with first term − 62 and d = 3. 1 1 t h term = − 62 + 10 × 3 = − 32 .
The 1 3 t h term from the end of the A.P. : 20, 13, 6, − 1 , ....., − 148 is :
(A) 57(B) − 57 (C) 64(D) − 64
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Answer: (D) − 64
Reverse the A.P.: − 148 , − 141 , − 134 , … with first term − 148 and d = 7. 1 3 t h term = − 148 + 12 × 7 = − 148 + 84 = − 64 .
The common difference of the A.P. whose n t h term is given by a n = 5 n − 7 is :
(A) − 7 (B) 7(C) 5(D) − 2
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Answer: (C) 5
a 1 = 5 − 7 = − 2 , a 2 = 10 − 7 = 3 .d = 3 − ( − 2 ) = 5 .
If the sum of the first n terms of an A.P be 3 n 2 + n and its common difference is 6, then its first term is
(A) 2(B) 3(C) 1(D) 4
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Answer: (D) 4
First term a 1 = S 1 = 3 ( 1 ) 2 + 1 = 4 . (Check: a 2 = S 2 − S 1 = 14 − 4 = 10 , so d = 6.)
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