Let a _ 1 , a _ 2 , a _ 3 , be an A.P. If ∑ _ r = 1 ^ ∞ a _ r 2 ^ r...

Let $S = \mathop{∑}_{r = 1}^{\infty}\dfrac{{a}_{r}}{{2}^{r}}$

$⇒ S = \dfrac{{a}_{1}}{2} + \dfrac{{a}_{2}}{{2}^{2}} + \dfrac{{a}_{3}}{{2}^{3}} + \ldots . . . . . .$\left(1\right)$</p><p>$⇒ \dfrac{S}{2} = \dfrac{{a}_{1}}{{2}^{2}} + \dfrac{{a}_{2}}{{2}^{3}} + \ldots . . . . . \left(2\right)$</p><p>Now subtracting both equation we get, </p><p>$⇒ \dfrac{S}{2} = \dfrac{{a}_{1}}{2} + d\left(\dfrac{1}{{2}^{2}} + \dfrac{1}{{2}^{3}} + \ldots\right)$</p><p>$⇒ \dfrac{S}{2} = \dfrac{{a}_{1}}{2} + d\left(\dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{2}}\right)$</p><p>$⇒ S = {a}_{1} + d = {a}_{2}$</p><p>Now given$S = \mathop{∑}_{r = 1}^{\infty}\dfrac{{a}_{r}}{{2}^{r}} = 4$</p><p>So,$S = {a}_{2} = 4$</p><p>So,$4{a}_{2} = 16