Let a_{1}, a_{2}, a_{3}, ldots be an A.P. If | vedclass

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(B) Let $S = \sum_{r=1}^{\infty} \frac{a_{r}}{2^{r}} = \frac{a_{1}}{2} + \frac{a_{2}}{2^{2}} + \frac{a_{3}}{2^{3}} + \ldots = 4$ Multiply by $\frac{1}

Vedclass · Accepted Answer

$16$

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(B) Let $S = \sum_{r=1}^{\infty} \frac{a_{r}}{2^{r}} = \frac{a_{1}}{2} + \frac{a_{2}}{2^{2}} + \frac{a_{3}}{2^{3}} + \ldots = 4$ Multiply by $\frac{1}{2}$: $\frac{S}{2} = \frac{a_{1}}{2^{2}} + \frac{a_{2}}{2^{3}} + \frac{a_{3}}{2^{4}} + \ldots$ Subtracting the two equations: $S - \frac{S}{2} = \frac{a_{1}}{2} + \frac{a_{2}-a_{1}}{2^{2}} + \frac{a_{3}-a_{2}}{2^{3}} + \ldots$ Since $a_{r}$ is an $A.P.$,$a_{r} - a_{r-1} = d$: $\frac{S}{2} = \frac{a_{1}}{2} + \frac{d}{2^{2}} + \frac{d}{2^{3}} + \frac{d}{2^{4}} + \ldots$ $\frac{S}{2} = \frac{a_{1}}{2} + d \left( \frac{1/4}{1 - 1/2} ight) = \frac{a_{1}}{2} + d \left( \frac{1/4}{1/2} ight) = \frac{a_{1}}{2} + \frac{d}{2}$ $S = a_{1} + d = a_{2}$ Given $S = 4$,so $a_{2} = 4$. Therefore,$4 a_{2} = 4 	imes 4 = 16$.

Let $a_{1}, a_{2}, a_{3}, \ldots$ be an $A.P.$ If $\sum_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4$,then $4 a_{2}$ is equal to

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