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

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

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$9$

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(C) Let the given series be $S = 3 + \frac{3+d}{4} + \frac{3+2d}{4^2} + \dots \infty$  --- $(1)$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 + \left( \frac{3+d}{4} - \frac{3}{4} \right) + \left( \frac{3+2d}{4^2} - \frac{3+d}{4^2} \right) + \dots \infty$$\frac{3S}{4} = 3 + \frac{d}{4} + \frac{d}{4^2} + \frac{d}{4^3} + \dots \infty$The series $3 + \frac{d}{4} + \frac{d}{4^2} + \dots$ is an Arithmetico-Geometric series where the sum of the infinite geometric part $\frac{d}{4} + \frac{d}{4^2} + \dots$ is $\frac{a}{1-r} = \frac{d/4}{1 - 1/4} = \frac{d/4}{3/4} = \frac{d}{3}$.So,$\frac{3S}{4} = 3 + \frac{d}{3}$.Given $S = 8$,substitute $S = 8$ into the equation:$\frac{3(8)}{4} = 3 + \frac{d}{3}$$6 = 3 + \frac{d}{3}$$3 = \frac{d}{3}$$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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