MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers

Answer: c
Explaination:Reason: We have an = 3 + 4n
∴ a_n+1 = 3 + 4(n + 1) = 7 + 4n
∴ d = a_n+1 – a_n
= (7 + 4n) – (3 + 4n)
= 7 – 3
= 4

Answer: c
Explaination:Reason: Since p, q, r, s are in A.P.
∴ (q – p) = (r – q) = (s – r) = d (common difference)

Answer: d
Explaination:Reason: Let three numbers be a – d, a, a + d
∴ a – d +a + a + d = 9
⇒ 3a = 9
⇒ a = 3
Also (a – d) . a . (a + d) = 24
⇒ (3 -d) .3(3 + d) = 24
⇒ 9 – d² = 8
⇒ d² = 9 – 8 = 1
∴ d = ± 1
Hence numbers are 2, 3, 4 or 4, 3, 2

Answer: d
Explaination:Reason: Here a = 7, d = 12-7 = 5
∴ a_n-1 = a + [(n – 1) – l]d = 7 + [(n – 1) -1] (5) = 7 + (n – 2)5 = 7 + 5n – 10 = 5M – 3

Answer: c
Explaination:Reason: Here a = 5, d = 2 – 5 = -3
a_n = a + (n – 1)d = 5 + (n – 1) (-3) = 5 – 3n + 3 = 8 – 3n

Answer: a
Explaination:Reason: Here l = -1000, d = -10 – (-5) = -10 + 5 = – 5
∴ 10^th term from the end = l – (n – 1 )d = -1000 – (10 – 1) (-5) = -1000 + 45 = -955

Answer: a
Explaination:Reason: Here a_n = 3n + 4
∴ a₁ = 7, a₂ – 10, a₃ = 13
∴ a= 7, d = 10 – 7 = 3
∴ S₁₂ = [2 × 7 + (12 – 1) ×3] = 6[14 + 33] = 6 × 47 = 282

Answer: b
Explaination:Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.
Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even
∴ Sum = [11 + 99] = × 110 = 45 × 55 = 2475

Answer: d
Explaination:Reason: Required Sum = 1 + 3 + 5 + … + upto n terms.
Here a = 1, d = 3 – 1 = 2
Sum = [2 × 1 + (n – 1) × 2] = [2 + 2n – 2] = × 2n = n²Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.
Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even
∴ Sum = [11 + 99] = × 110 = 45 × 55 = 2475

Answer: d
Explaination:Reason: Let a is first term and d is common difference
∴ a_{p + q} = m
a_{p – q} = n
⇒ a + (p + q – 1)d = m = …(i)
⇒ a + (p – q – 1)d = m = …(ii)
On adding (i) and (if), we get
2a + (2p – 2)d = m + n
⇒ a + (p -1)d = …[Dividing by 2
∴ a_n =

Answer: a
Explaination:Reason: Since a, b, c are in A.P.
∴ b – a = c – b
⇒ = 1
⇒ = 1

Answer: d
Explaination:Reason: Multiples of n from 1 to n² are n × 1, n × 2, n × 3, …, m× n
∴ There are n numbers
Thus, the number of mutiples of n which lie between n and n² is (n – 2) leaving first and last in the given list: Total numbers are (n – 2).

Answer: a
Explaination:Reason: Let common difference of A.P. be x
∴ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x
Given equation n-4b + 6c-4d + c
= a – 4(a + x) + 6(A + 2r) – 4(n + 3x) + (o + 4.v)
= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x = 8a – 8a + 16x – 16x = 0

Answer: a
Explaination:

Answer: a
Explaination:Reason: an = a + (n – 1)d

Answer: a
Explaination:Reason: Here l – 254, d = 9-4 = 5
∴ 10^th term from the end = l – (10 – 1 )d = 254 -9d = 254 = 9(5) = 254 – 45 = 209

Answer: d
Explaination:Reason: Since 2x, x + 10 and 3x + 2 are in A.P.
∴ 2(x + 10) = 2x + (3x + 2)
⇒ 2x + 20 – 5x + 2
⇒ 2x – 5x = 2 – 20
⇒ 3x = 18
⇒ x = 6

Answer: b
Explaination:Reason: The numbers are 3, 9,15, 21, …, 99
Here a = 3, d = 6 and a_n = 99
∴ a_n = a + (n – 1 )d
⇒ 99 = 3 + (n – 1) x 6
⇒ 99 = 3 + 6n – 6
⇒ 6n = 102
⇒ n = 17
Required Sum = [a + a_n] = [3 + 99] = × 102 = 867

Answer: a
Explaination:Reason: Let x be the common difference of the given AP
∴ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x
∴ a – 4b + 6c – 4d + e = a – 4 (a + x) + 6(a + 2x) – 4(a + 3x) + (a + 4x)
= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x = 8a – 8a + 16x – 16x = 0

Answer: d
Explaination:Reason: We have 7a₇ = 11a₁₁
⇒ 7[a + (7 – 1)d] = 11[a + (11 – 1 )d]
⇒ 7(a + 6d) = 11(a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 4a = -68d
⇒ a = -17d
∴ a₁₈ = a + (18 – 1)d = a + 17d = -17d + 17d = 0