Mathematics · 2019
JEE Main · 10 April 2019, Shift 2 · Q71
Let a_1, a_2, a_3, ….. be an A.P. with a_6=2. Then the common difference of this A.P., which maximises the product a_1 a_4 a_5, is:
Let $\displaystyle \mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots .$. be an A.P. with $\displaystyle \mathrm{a}_6=2$. Then the common difference of this A.P., which maximises the product $\displaystyle \mathrm{a}_1 \mathrm{a}_4 \mathrm{a}_5$, is :
Official answer
(2)
$\displaystyle \frac{8}{5}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.