JEE Main 2025 Jan 24 Shift 2, Mathematics Q10: Summation of Series

Split the series into two infinite geometric sums.

The "5" part:

$5\left(1+\frac{1}{7}+\frac{1}{7^2}+\dots \right)=5\cdot \frac{1}{1-1/7}=5\cdot \frac{7}{6}=\frac{35}{6}$

The $\alpha$ part is an arithmetic-geometric-like series, which is most cleanly handled as a derivative of a geometric series. Equivalently, multiply the whole right side $S=7$ by $\frac{1}{7}$ and subtract.

$S=5+\frac{5+\alpha }{7}+\frac{5+2\alpha }{7^2}+\frac{5+3\alpha }{7^3}+\dots$

$\frac{S}{7}=\frac{5}{7}+\frac{5+\alpha }{7^2}+\frac{5+2\alpha }{7^3}+\dots$

Subtract: $\frac{6S}{7}=5+\frac{\alpha }{7}+\frac{\alpha }{7^2}+\frac{\alpha }{7^3}+\dots =5+\alpha \cdot \frac{1/7}{1-1/7}=5+\frac{\alpha }{6}$

Using $S=7$: $\frac{6\cdot 7}{7}=6=5+\frac{\alpha }{6}\Rightarrow \frac{\alpha }{6}=1\Rightarrow \alpha =6$.

Correct option: (3)