If 3 + frac{1}{4} (3 + d) + frac{1}{4^2} (3 + | vedclass

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(C) Let $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$  $(1)$Multiply by $\frac{1}{4}$:$\frac{S}{4} = \frac{3}{4} + \frac{3+d}{4^2} + \frac

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$9$

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(C) Let $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$  $(1)$Multiply by $\frac{1}{4}$:$\frac{S}{4} = \frac{3}{4} + \frac{3+d}{4^2} + \frac{3+2d}{4^3} + \dots \infty$  $(2)$Subtract $(2)$ from $(1)$:$S - \frac{S}{4} = 3 + \frac{d}{4} + \frac{d}{4^2} + \frac{d}{4^3} + \dots \infty$$\frac{3S}{4} = 3 + \frac{d/4}{1 - 1/4} = 3 + \frac{d/4}{3/4} = 3 + \frac{d}{3}$Given $S = 8$,so $\frac{3(8)}{4} = 3 + \frac{d}{3}$$6 = 3 + \frac{d}{3}$ $\Rightarrow 3 = \frac{d}{3}$ $\Rightarrow d = 9$

If $3 + \frac{1}{4} (3 + d) + \frac{1}{4^2} (3 + 2d) + \dots \infty = 8$,then the value of $d$ is:

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