If 8 = 3 + frac{1}{4}(3 + p) + frac{1}{4^2}(3 | vedclass

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(A) The given series is an Arithmetico-Geometric Progression $(AGP)$ of the form $\sum_{n=0}^{\infty} (a + np)r^n$,where $a = 3$,$d = p$,and $r = \fra

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$9$

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(A) The given series is an Arithmetico-Geometric Progression $(AGP)$ of the form $\sum_{n=0}^{\infty} (a + np)r^n$,where $a = 3$,$d = p$,and $r = \frac{1}{4}$.The sum of an infinite $AGP$ is given by $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$.Substituting the values into the formula:$8 = \frac{3}{1 - \frac{1}{4}} + \frac{p \cdot \frac{1}{4}}{(1 - \frac{1}{4})^2}$$8 = \frac{3}{\frac{3}{4}} + \frac{\frac{p}{4}}{(\frac{3}{4})^2}$$8 = 4 + \frac{\frac{p}{4}}{\frac{9}{16}}$$8 - 4 = \frac{p}{4} \cdot \frac{16}{9}$$4 = \frac{4p}{9}$$p = 9$.

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

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