Mathematics · 2013
JEE Main · 9 April 2013, Shift 1 · Q68
Let a_1, a_2, a_3, … be an A.P. such that (a_1+ a_2+…+ a_p)/(a_1+ a_2+ a_3+…+ a_q)=(p^3)/(q^3); p ≠ q. Then (a_6)/(a_21) is equal to:
Let $\displaystyle a_1, a_2, a_3, \ldots$ be an A.P. such that $\displaystyle \frac{\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{\mathrm{p}}}{\mathrm{a}_1+\mathrm{a}_2+\mathrm{a}_3+\ldots+\mathrm{a}_{\mathrm{q}}}=\frac{\mathrm{p}^3}{\mathrm{q}^3} ; \mathrm{p} \neq \mathrm{q}$. Then $\displaystyle \frac{\mathrm{a}_6}{\mathrm{a}_{21}}$ is equal to :
Official answer
(2)
$\displaystyle \frac{121}{1681}$
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JEE Main 2013 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.