Sum of first 13 terms of a GP is equal to the sum of the first 11 terms in the same GP. Sum of the first 15 terms is 1200, what is the $2 1^{s t}$ term in the same GP?

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UP Police SI (दरोगा) Official PYP (Held On: 24 Nov 2021 Shift 1)

Answer 1

Given:

Sum of first 13 terms of a GP = Sum of first 11 terms of the same GP

Sum of first 15 terms of the GP = 1200

Formula used:

Sum of n terms of a GP: S_n = a(1 - rⁿ) / (1 - r)

To find the 21st term: T₂₁ = a × r²⁰

Calculation:

⇒ S₁₃ = S₁₁

⇒ a(1 - r¹³) / (1 - r) = a(1 - r¹¹) / (1 - r)

⇒ 1 - r¹³ = 1 - r¹¹

⇒ r¹³ = r¹¹

⇒ r² = 1

⇒ r = ±1

⇒ r = -1 (since r = 1 not valid for equal sums)

Now using S₁₅ = 1200

⇒ a(1 - (-1)¹⁵) / (1 - (-1)) = 1200

⇒ a(1 + 1) / 2 = 1200

⇒ 2a / 2 = 1200

⇒ a = 1200

T₂₁ = a × r²⁰

⇒ T₂₁ = 1200 × (-1)²⁰

⇒ T₂₁ = 1200 × 1

⇒ T₂₁ = 1200

∴ The correct answer is option (3).

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