Mathematics · 2022
JEE Main · 29 July 2022, Shift 1 · Q4
If 1/((20-a)(40-a))+1/((40-a)(60-a))+…+1/((180-a)(200-a))=1/256, then the maximum value of a is:
If $\displaystyle \frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$, then the maximum value of a is :
Official answer
(3)
$\displaystyle 212$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.