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 = \(\frac{12}{2}\)[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 = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 55 = 2475

Answer: d
Explaination:Reason: Required Sum = 1 + 3 + 5 + … + upto n terms.
Here a = 1, d = 3 – 1 = 2
Sum = \(\frac{n}{2}\)[2 × 1 + (n – 1) × 2] = \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\) × 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 = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 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 = \(\frac{m+n}{2}\) …[Dividing by 2
∴ a_n = \(\frac{m+n}{2}\)

Answer: a
Explaination:Reason: Since a, b, c are in A.P.
∴ b – a = c – b
⇒ \(\frac{b-a}{c-b}\) = 1
⇒ \(\frac{a-b}{b-c}\) = 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 = \(\frac{n}{2}\)[a + a_n] = \(\frac{17}{2}\)[3 + 99] = \(\frac{17}{2}\) × 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 =\(\frac{31}{2}\) (2a + 30d)
a₁₆ = a + 15d = zw
⇒ S₃₁ = \(\frac{31}{2}\) × 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:

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 = \(\frac{-21}{-3}\)
⇒ 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

Answer:
Explaination:
∵ a = p and d = q
a_n = a + (n – 1)d
⇒ a₁₀ =p + 9q

Answer:
Explaination:
Here, a = 14, d= 11 – 14 = – 3
Let a_n = – 1
⇒ a + (n – 1 )d = – 1
⇒ 14 + (n – 1)(- 3) = – 1
⇒ (n – 1)(- 3) = – 1 – 14
⇒ n – 1 = \(\frac{-15}{-3}\) = 5
⇒ n = 6
∴ 6th term of the AP is -1.

Answer:
Explaination:
a₁ = 1
a₁ = a₂ – a₁ = – 1 – 1 = – 2
a₅ = a₁ + 4d
= 1 + 4(-2)
= 1 – 8 = — 7
a₆ = a₅ + d = – 7 – 2 = – 9
Next two terms are – 7 and – 9.

Answer:
Explaination:

Answer:
Explaination:
S₄₀(2 × 2 + 39 × 4)
= 40 × (80)= 3200

Answer:
Explaination:
Let numbers be a – d, a and a + d.
⇒ a – d+a + a + d= 24
⇒ 3a = 24
⇒ a = 8
∴ Middle term = 8.

Answer:
Explaination:
1 – 6 + 2 – 7 + 3 – 8 + … 100 terms
= (1 + 2 + 3 +… up to 50 terms) + (-6 -7 – 8 – … up to 50 terms)
= \(\frac{50}{2}\)[2 × 1 + 49 × 1] + [\(\frac{50}{2}\){2 × (-6) + 49 × (-1)}] = 25 × 51 + 25 x (-12 – 49) = 25(51 – 61)
= -250.

Answer:
Explaination:
S_{m = 2m² + 3m
∴ S1 = 2 × 1² + 3 × 1 = 5 = a1
and S2 = 2 × 2² + 3 × 2 = 14 …(i)
⇒ a1 + a2 = 14
⇒ 5 + a2 = 14
⇒ a2 = 9}

Answer:
Explaination:
S_{p = ap² + bp
S1= a × 1² + b × 1
= a + b = a1
and S2 = a × 2² + b × 2 = 4a + 2b …(i)
⇒ a1 + a2 = 4a + 2b
⇒ a + b + a2 = 4a + 2b
⇒ a2 = 3a + b [Using eq. (i)] Now, d = a2 – a1
= (3a + b)-(a + b) = 2a.}

Answer:
Explaination:
S_{n = 2n² + 5n
S20 = 2(20)² + 5 × 20
= 2 × 400 + 100 = 900}