Mathematics · 2026
JEE Main · 24 January 2026, Shift 2 · Q6
(1/3+4/7)+(1/3^2+1/3 × 4/7+4^2/7^2)+(1/3^3+1/3^2 × 4/7+1/3 × 4^2/7^2+4^3/7^3)+… upto infinite terms, is equal to
$\displaystyle \left(\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{3^2}+\frac{1}{3} \times \frac{4}{7}+\frac{4^2}{7^2}\right)+\left(\frac{1}{3^3}+\frac{1}{3^2} \times \frac{4}{7}+\frac{1}{3} \times \frac{4^2}{7^2}+\frac{4^3}{7^3}\right)+\ldots$ upto infinite terms, is equal to
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \frac{5}{2}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.