Mathematics - Sequence & Series Question with Solution | TestHub

Let $S_n$ be the sum of the first $n$ terms of the G.P.

Given $\frac{S_3}{S_6}=\frac{64}{91}$, we have $\frac{a(1-r^3)}{a(1-r^6)}=\frac{64}{91}$.

Then $\frac{1}{1+r^3}=\frac{64}{91}$, so $1+r^3=\frac{91}{64}$, $r^3=\frac{27}{64}$. Thus, $r=\frac{3}{4}$.

Final Answer: The final answer is $\boxed{\frac{3}{4}}$