Mathematics · 2021
JEE Main · 31 August 2021, Shift 2 · Q65
Let a_1, a_2, a_3, … be an A.P. If (a_1+ a_2+…+ a_10)/(a_1+ a_2+…+ a_p)=100/(p^2), p ≠ 10, then (a_11)/(a_10) is equal to:
Let $\displaystyle \mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots$ be an A.P. If $\displaystyle \frac{\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}}{\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{\mathrm{p}}}=\frac{100}{\mathrm{p}^2}, \mathrm{p} \neq 10$, then $\displaystyle \frac{\mathrm{a}_{11}}{\mathrm{a}_{10}}$ is equal to :
Official answer
(2)
$\displaystyle \frac{21}{19}$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.