Mathematics · 2023
JEE Main · 30 January 2023, Shift 1 · Q66
If a_n =-2/(4 n^2-16 n +15), then a_1+ a_2+… …+ a_25 is equal to:
If $\displaystyle \mathrm{a}_{\mathrm{n}}=\frac{-2}{4 \mathrm{n}^2-16 \mathrm{n}+15}$, then $\displaystyle \mathrm{a}_1+\mathrm{a}_2+\ldots \ldots+\mathrm{a}_{25}$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{50}{141}$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.