If 7 = 5 + frac{1}{7}(5 + alpha) + frac{1}{7^ | vedclass

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(C) Let $S = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \dots \infty$. Given $S = 7$. Multiply by $\frac{1}{7}$: $\frac{1}{7}S = \frac

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$6$

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(C) Let $S = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \dots \infty$. Given $S = 7$. Multiply by $\frac{1}{7}$: $\frac{1}{7}S = \frac{1}{7}(5) + \frac{1}{7^2}(5 + \alpha) + \frac{1}{7^3}(5 + 2\alpha) + \dots \infty$. Subtracting the two equations: $S - \frac{1}{7}S = 5 + [\frac{1}{7}(5 + \alpha - 5) + \frac{1}{7^2}(5 + 2\alpha - (5 + \alpha)) + \dots \infty]$. $\frac{6}{7}S = 5 + [\frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots \infty]$. The term in the bracket is an infinite geometric series with first term $a = \frac{\alpha}{7}$ and common ratio $r = \frac{1}{7}$. Sum $= \frac{a}{1-r} = \frac{\alpha/7}{1 - 1/7} = \frac{\alpha/7}{6/7} = \frac{\alpha}{6}$. So,$\frac{6}{7}S = 5 + \frac{\alpha}{6}$. Since $S = 7$,we have $\frac{6}{7}(7) = 5 + \frac{\alpha}{6}$. $6 = 5 + \frac{\alpha}{6}$ $\Rightarrow 1 = \frac{\alpha}{6}$ $\Rightarrow \alpha = 6$.

If $7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \frac{1}{7^3}(5 + 3\alpha) + \dots \infty$,then the value of $\alpha$ is:

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