JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Sequences and Series

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The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

(A) $\frac{34}{9}$ ✓ Correct
(B) $\frac{34}{13}$
(C) $\frac{32}{9}$
(D) $\frac{32}{13}$

Answer: (A)

Solution

For the A.P.,

S_{10} = 5 (2 a + 9 d) = 160

, so

2 a + 9 d = 32

. The G.P. has first term

d

and common ratio

a

, so

d + d a = 8

, i.e.

d (1 + a) = 8

. Eliminating

a = \frac{32 - 9 d}{2}

gives

d \cdot \frac{34 - 9 d}{2} = 8

, i.e.

9 d^{2} - 34 d + 16 = 0

. The two possible first terms of the G.P. are the roots

d

, whose sum is

\frac{34}{9}

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

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