Question:

Step 1: For $\sin x$, $(\sin x + 1)$, and $6(\sin x + 1)$ to be in a Geometric Progression (G.P.), the square of the middle term must equal the product of the first and third terms. $$(\sin x + 1)^2 = \sin x \cdot 6(\sin x + 1)$$ Since $0 < x < \pi/2$, we know that $\sin x > 0$, which implies $\sin x + 1 > 1$. Therefore, $\sin x + 1 \neq 0$, and we can divide both sides by $(\sin x + 1)$: $$\sin x + 1 = 6\sin x$$ Subtracting $\sin x$ from both sides yields: $$1 = 5\sin x$$ Solving for $\sin x$: $$\sin x = \frac{1}{5}$$ Step 2: Determine the common ratio $r$ of the G.P. The common ratio is the ratio of any term to its preceding term. Using the first two terms: $$r = \frac{\sin x + 1}{\sin x}$$ Substitute the value $\sin x = \frac{1}{5}$: $$r = \frac{\frac{1}{5} + 1}{\frac{1}{5}} = \frac{\frac{6}{5}}{\frac{1}{5}} = 6$$ Alternatively, using the second and third terms: $$r = \frac{6(\sin x + 1)}{\sin x + 1} = 6$$ Thus, the common ratio of the G.P. is $6$.