Question

Let $a_1,a_2,a_3,\dots$ be an A.P. If $\sum _{r=1}^{\infty }\frac{a_r}{2^r}=4$, then $4a_2$ is equal to

Answer 1

Answer: 16

$S=\frac{a_1}{2}+\frac{a_2}{2^2}+\frac{a_3}{2^3}+\dots$
$\frac{\frac{S}{2}=\frac{a_1}{2^2}+\frac{a_2}{2^3}+\dots }{\frac{S}{2}=\frac{a_1}{2}+d\left(\frac{1}{2^2}+\frac{1}{2^3}+\dots \right)}$
$\frac{S}{2}=\frac{a_1}{2}+d\left(\frac{\frac{1}{4}}{1-\frac{1}{2}}\right)$
$\therefore S=a_1+d=a_2=4$
Or $4a_2=16$