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  4. Let a1, a2, a3, ldots be an A.P. If displaystyle∑r=1∞ (ar/2r)=4, then 4 a2 is equal to

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

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Solution:

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