MCQ
If $7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^{2}}(5+2 \alpha)+\frac{1}{7^{3}}(5+3 \alpha)+$ $\infty$, then the value of $\alpha$ is:
  • A
    1
  • B
    $\frac{6}{7}$
  • 6
  • D
    $\frac{1}{7}$

Answer

Correct option: C.
6
(C)
Sol. Let $\mathrm{S}=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^{2}}(5+2 \alpha)+\ldots$
$\frac{1}{7} \mathrm{~S}=\frac{1}{7}(5)+\frac{1}{7^{2}}(5+\alpha)+\ldots \infty$
$\frac{6}{7}(S)=5+\frac{1}{7} \alpha\left(\frac{1}{1-\frac{1}{7}}\right)$
$6=5+\frac{\alpha}{6} \Rightarrow \alpha=6$

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