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

JEE · Math · previous-year question

  1. A.$\frac{1}{7}$
  2. B.1
  3. C.$\frac{6}{7}$
  4. D.6correct

Answer

D. 6

Explanation

$$\begin{aligned} & \text { Let } S=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2 \alpha)+\ldots \\ & \frac{1}{7} S=\frac{1}{7}(5)+\frac{1}{7^2}(5+\alpha)+\ldots \infty \end{aligned}$$ $$\begin{aligned} &\frac{6}{7}(S)=5+\frac{1}{7} \alpha\left(\frac{1}{1-\frac{1}{7}}\right)\\ &6=5+\frac{\alpha}{6} \Rightarrow \alpha=6 \end{aligned}$$

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