Let a 1 , frac a 2 2 , frac a 3 2 2 , ldots, frac a 10 2 9 be a G.P. of common ratio frac 1 √(2) . If a 1 + a 2 +ldots+ a 10 =62 , then a 1 is equal to :

Mathematics · Sequence and Series

JEE Main 2026 — 21 January, Evening Shift — Question 12

Let $\mathrm{a}_{1}, \frac{\mathrm{a}_{2}}{2}, \frac{\mathrm{a}_{3}}{2^{2}}, \ldots, \frac{\mathrm{a}_{10}}{2^{9}}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$ . If $\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{10}=62$ , then $\mathrm{a}_{1}$ is equal to :

Option A:
$2(\sqrt{2}-1)$
Correct
Option B:
$2-\sqrt{2}$
Option C:
$\sqrt{2}-1$
Option D:
$2(2-\sqrt{2})$

Answer: A

Step-by-step solution

\text{Given G.P.: } & T_n = \frac{a_n}{2^{n-1}} \text{ with } r = \frac{1}{\sqrt{2}} \\ \implies \frac{a_n}{2^{n-1}} &= a_1 \left( \frac{1}{\sqrt{2}} \right)^{n-1} \\ a_n &= a_1 \cdot \left( \frac{2}{\sqrt{2}} \right)^{n-1} = a_1 (\sqrt{2})^{n-1} \\ \text{Sum } S_{10} &= a_1 + a_2 + \dots + a_{10} = 62 \\ 62 &= \frac{a_1 ((\sqrt{2})^{10} - 1)}{\sqrt{2} - 1} \\ 62 &= \frac{a_1 (32 - 1)}{\sqrt{2} - 1} \\ 2 &= \frac{a_1}{\sqrt{2} - 1} \implies a_1 = 2(\sqrt{2}-1) \end{aligned}$$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Geometric Progression