Mathematics - Sequence & Series Question with Solution | TestHub

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

So, on simplifying we get,

$\Rightarrow \frac{1}{20}\left[\left(\frac{1}{20-a}-\frac{1}{40-a}\right)+\left(\frac{1}{40-a}-\frac{1}{60-a}\right)\right......$

$\left.+\dots +\left(\frac{1}{180-a}-\frac{1}{200-a}\right)\right]=\frac{1}{256}$

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

$\Rightarrow \left(20-a\right)\left(200-a\right)=256\times 9$

$\Rightarrow a^2-220a+1696=0$

$\Rightarrow a=8,212$

So, maximum value of $a$ is $212$.