If 1(20-a)(40-a)+1(40-a)(60-a)++1(180-a)(200-a)=1256, then the maximum value of a is:

Question

If $\dfrac{1}{(20-a)(40-a)}+\dfrac{1}{(40-a)(60-a)}+\cdots+\dfrac{1}{(180-a)(200-a)}=\dfrac{1}{256}$, then the maximum value of $a$ is:

Accepted Answer

$${1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {40 - a}} + {1 \over {40 - a}} - {1 \over {60 - a}}\, + \,....\, + \,{1 \over {180 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$

$$ \Rightarrow {1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$

$$ \Rightarrow {1 \over {20}}\left( {{{180} \over {(20 - a)(200 - a)}}} \right) = {1 \over {256}}$$

$$ \Rightarrow (20 - a)(200 - a) = 9.256$$

OR ${a^2} - 220a + 1696 = 0$

$$ \Rightarrow a = 212,\,8$$

Solutions · 1