JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 4)
If $$\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$$, then the maximum value of $$\mathrm{a}$$ is :
198
202
212
218
Explanation
$${1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {40 - a}} + {1 \over {40 - a}} - {1 \over {60 - a}}\, + \,....\, + \,{1 \over {180 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$
$$ \Rightarrow {1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$
$$ \Rightarrow {1 \over {20}}\left( {{{180} \over {(20 - a)(200 - a)}}} \right) = {1 \over {256}}$$
$$ \Rightarrow (20 - a)(200 - a) = 9.256$$
OR $${a^2} - 220a + 1696 = 0$$
$$ \Rightarrow a = 212,\,8$$
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