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: 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

Answer: b
Explaination:
∵ p, q, r are in AP.
∴ 2q = p + r
⇒ p + r - 2q = 0
∴ p³ + r³ + (-2p)³ = 3 × p × r × -2q
[Using ifa + 6 + c = 0 ⇒ a³ + b³ + c³ = 3 abc]
⇒ p³ + r³ - 8q³ = -6pqr.

Answer: b
Explaination: (b) a₁₀₁ = 3.5 + 0(100) = 3.5

Answer: b
Explaination: (b) An AP with d = 4.

Answer: c
Explaination:
a₄ - b₄ = (a₁ + 3d) - (b₁ + 3d)
= a₁ - b₁= - 1 - (-8) = 7

Answer: d
Explaination:
d = - 2, n = 5, a_n = 0
∵ a_n = 0
⇒ a + (n - 1)d=0
⇒ a + (5 - 1)(- 2) = 0
⇒ a = 8
Correct option is (d).

Answer: c
Explaination:
Common difference, d = 3
a₂₀ - a₁₅ = (a + 19d) - (a+ 14d)
= 5d=5 × 3 = 15

Answer: c
Explaination:
(c) √18, √50, √98, ..... = 3√2, 5√2, 7√2, ......
∴ Next term is 9√2 = √162

Answer: c
Explaination: (c) Common difference = a₂ - a₁

Answer: b
Explaination:
a₁= 2 × 1 + 1 = 3,
a₂ = 2 × 2 + 1 = 5,
a₃ = 2 × 3 + l= 7
∴ Sum = 3 + 5 + 7 = 15

Answer: c
Explaination:
S₃₁ = (2a + 30d)
a₁₆ = a + 15d = zw
⇒ S₃₁ = × 2(a + 15d)
⇒ S₃₁ = 31m

Answer: d
Explaination:

Answer: d
Explaination:
S_n = An + Bn²
S₁ = A × 1 +B × 1²
= A + B
∵ S₁ = a₁
∴ a₁ = A + B ... (1)
and S₂ = A × 2 + B × 2²
⇒ a₁ + a₂ = 2A + 4B
⇒ (A + B) + a₂ = 2A + 4B[Using (i)]
⇒ a₂ = A + 3B
∴ d = a₂ - a₁ = 2B

Answer:
Explaination:
∵ p - 1, p + 3 and 3p - 1 are in AP.
∴ 2(p + 3) = p - 1 + 3p - 1
⇒ 2p + 6 = 4/> -2.
⇒ -2p = -8
⇒ p = 4.

Answer:
Explaination:
(i) T_n= 2n + 3
T₁ = 2 × 1 + 3 = 5,
T₂ = 2 × 2 + 3 = 7,
T₃ = 2 × 3 + 3 = 9,
T₄ = 2 × 4 + 3 = 11
∴ Ist four terms are 5, 7, 9 and 11.

(ii) T_n = 3ⁿ⁺¹
⇒ T₁ = 3¹⁺¹ = 9,
T₂ = 3²⁺¹ = 27,
T₃= 3³⁺¹ = 81,
T₄ = 3⁴⁺¹ = 243
∴ Ist four terms are 9, 27, 81 and 243

(iii) T₁ = 2, T_n = T_n-1 + 5, n ≥ 2
⇒ T₂ = T_2-1 + 5
= T₁ + 5 = 2 + 5 = 7
T₃ = T_3-1 + 5
= T₂ + 5 = 7 + 5 = 12
and T₄=T_4-1 + 5
= T₃ + 5 = 12 + 5 = 17
∴ Ist four terms are 2, 7, 12 and 17.

Answer:
Explaination:
a=10, d = 10.5 - 10 = 0.5
a₁₀ = a + 9d= 10 + 9 × 0.5 = 14.5

Answer:
Explaination:
Let a = first term,
Given, d = - 4, a₇ = 4
⇒ a + (7 - 1)d = 4
[ ∵ nth term of an AP = a_n = a + (n - 1)d]
⇒ a + 6d = 4
⇒ a + 6 × (-4) = 4
⇒ a - 24 = 4.
⇒ a = 4 + 24 = 28
∴ First term, a = 28

Answer:
Explaination:
Here, a = 21, J= 18-21 = -3
Let a_n = 0
⇒ a + (n - 1 )d = 0
⇒ 21 +(n - 1)(-3) = 0
⇒ (n - 1)(- 3) = -21
⇒ n - 1 =
⇒ n = 8
∴ 8th term is zero.

Answer:
Explaination:
If terms are in AP, then
13 - (2p + 1) = (5p - 3) - 13
⇒ 13 - 2p - 1 = 5p - 3 - 13
⇒ 28 = 7p
⇒ p = 4.

Answer:
Explaination:
Let ‘cf be the common difference of the AP whose first term is ‘a’
Now, a₂₁ - a₇ = 84
⇒ (a + 20d) - (a + 6d)= 84
⇒ 20d - 6d = 84
⇒ 14d = 84
⇒ d = 6