Mathematics · 2026
JEE Main · 21 January 2026, Shift 2 · Q7
Let a_1, (a_2)/2, (a_3)/2^2, …, (a_10)/2^9 be a G.P. of common ratio 1/(√ 2). If a_1+ a_2+…+ a_10=62, then a_1 is equal to:
Let $\displaystyle \mathrm{a}_1, \frac{\mathrm{a}_2}{2}, \frac{\mathrm{a}_3}{2^2}, \ldots, \frac{\mathrm{a}_{10}}{2^9}$ be a G.P. of common ratio $\displaystyle \frac{1}{\sqrt{2}}$. If $\displaystyle \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=62$, then $\displaystyle \mathrm{a}_1$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 2(\sqrt{2}-1)$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.