If $7 = 5+\dfrac{1}{7}(5+\alpha)+\dfrac{1}{7^2}(5+2\alpha)+\cdots$ to infinity, then $\alpha$ equals
Step-by-Step Solution
Key Concept: Recognise this as an AGP with first term 5, common ratio $r=1/7$, and arithmetic increment $\alpha$; apply $S=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}$.
AGP with $a=5$, $d=\alpha$, $r=\frac{1}{7}$:
$$S = \frac{a}{1-r}+\frac{dr}{(1-r)^2} = \frac{5}{6/7}+\frac{\alpha\cdot(1/7)}{(6/7)^2} = \frac{35}{6}+\frac{\alpha/7}{36/49} = \frac{35}{6}+\frac{7\alpha}{36}.$$
Setting $S=7$: $\dfrac{7\alpha}{36}=7-\dfrac{35}{6}=\dfrac{42-35}{6}=\dfrac{7}{6}\Rightarrow\alpha=6$.
Correct Answer: 2