Question

An arithmetic progressions in which first term is 4. Also, the sum of the first ten terms is 25/56 times of the sum of the first sixteen terms. Then, find the sum of the first 52 terms?
1. 1972
2. 1976
3. 1964
4. 1986

Answer 1

Correct Answer - Option 2 : 1976

Given:

a = 4,

S₁₀ = 25/56 × S₁₆

Formula used:

S_n = n/2 × [2a + (n – 1) d]

Where, a = first term, n = number of terms and d = common difference

Calculation:

S₁₀ = 10/2 × [2(4) + (10 – 1) d]

⇒ S₁₀ = 5(8 + 9d)

⇒ S₁₀ = 40 + 45d     ---- (1)

S₁₆ = 16/2 × [2(4) + (16 – 1) d]

⇒ S₁₆ = 8 × (8 + 15d)

⇒ S₁₆ = 64 + 120d     ---- (2)

According to question –

⇒ S₁₀ = 25/56 × S₁₆

⇒ 40 + 45d = 25/56 × (64 + 120d)

⇒ 2240 + 2520d = 1600 + 3000d

⇒ (3000d – 2520d) = (2240 – 1600)

⇒ 480d = 640

⇒ d = 640/480

⇒ d = 4/3

Now, S₅₂ = 52/2 × [2(4) + (52 – 1) × 4/3]

⇒ S₅₂ = 26[8 + 68]

⇒ S₅₂ = 26 × 76

∴ S₅₂ = 1976