$$7=5+\dfrac{1}{7}(5+\alpha)+\dfrac{1}{7^2}(5+2\alpha)+\dfrac{1}{7^3}(5+3\alpha)+\cdots+\infty$$
$$7=5\left(1+\dfrac{1}{7}+\dfrac{1}{7^2}+\cdots+\infty\right)+\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$
$$1+\dfrac{1}{7}+\dfrac{1}{7^2}+\cdots+\infty$$ is an infinite G.P with the common ratio of $$\dfrac{1}{7}$$.
Sum of the above G.P = $$\dfrac{1}{1-\dfrac{1}{7}}=\dfrac{7}{6}$$
$$7=\left(5\times\dfrac{7}{6}\right)+\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$
$$7-\dfrac{35}{6}=\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$
$$\dfrac{7}{6}=\alpha\ \left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ \right)$$
Now, $$S=\left(\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\right)$$ in the form of AGP where $$\left(1,\ 2,\ 3,\ \cdots,\ \infty\right)$$ in A.P and $$\left(\dfrac{1}{7},\ \dfrac{1}{7^2},\ \dfrac{1}{7^3},\ \cdots,\ \infty\right)$$ are in G.P.
$$S=\dfrac{1}{7}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\cdots+\infty\ $$ $$\longrightarrow\ i$$
Multiply both sides by $$\dfrac{1}{7}$$.
$$\dfrac{S}{7}=\dfrac{1}{7^2}+\dfrac{2}{7^3}+\dfrac{3}{7^4}+\cdots+\infty\ $$ $$\longrightarrow\ ii$$
Subtract equation $$ii$$ from equation $$i$$,
$$\dfrac{6S}{7}=\dfrac{1}{7}+\left(\dfrac{2}{7^2}-\dfrac{1}{7^2}\right)+\left(\dfrac{3}{7^3}-\dfrac{2}{7^3}\right)+\left(\dfrac{4}{7^4}-\dfrac{3}{7^4}\right)+\cdots+\infty$$
$$\dfrac{6S}{7}=\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+\dfrac{1}{7^4}+\cdots+\infty$$
$$\dfrac{6S}{7}=\dfrac{\dfrac{1}{7}}{\left(1-\dfrac{1}{7}\right)}$$
$$\dfrac{6S}{7}=\dfrac{1}{6}$$
$$S=\dfrac{7}{36}$$
Hence, $$\dfrac{7}{6}=\alpha\ \times\dfrac{7}{36}$$
$$\alpha\ =6$$
Hence, the value of $$\alpha$$ is 6.
$$\therefore\ $$ The required answer is B.
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