JEE Math Practice Question
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 :
- A.$\frac{1}{7}$
- B.1
- C.$\frac{6}{7}$
- D.6
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