Animated Solution for Mathematics - Sequence and Series: If $7=5+\frac{1}{7}(5+\alpha )+\frac{1}{7^2}(5+2\alpha )+\frac{1}{7^3}(5+3\alpha )+\dots \infty$, then the value of $\alpha$ is :

We are examining the infinite series:

The numerators follow an arithmetic progression with a common difference of $\alpha$, while the denominators follow a geometric progression with a common ratio of $r=\frac{1}{7}$. This structure identifies the series as an Arithmetico-Geometric Progression (AGP).

The Art of Shifting

To solve this, we employ the "shift and subtract" method. We multiply the entire series $S$ by the common ratio $r=\frac{1}{7}$:

Now, we subtract this shifted series from the original series $S$. By aligning terms with identical denominators, we obtain:

The terms within the parentheses simplify significantly as the constant $5$ cancels out in each pair. This leaves us with:

The terms following the initial $5$ constitute an infinite geometric progression with the first term $a=\frac{\alpha }{7}$ and common ratio $r=\frac{1}{7}$. Using the sum formula for an infinite GP, $S_{\infty }=\frac{a}{1-r}$, we get:

Simplifying the denominator $1-\frac{1}{7}=\frac{6}{7}$, the equation becomes:

Given that the total sum $S=7$, we substitute this value into the equation:

Subtracting $5$ from both sides yields $1=\frac{\alpha }{6}$. Therefore, the final value is: