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:
- A1
- B$\frac{6}{7}$
- ✓6
- D$\frac{1}{7}$
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