In a G.P. the ratio of the sum of first 3 terms is to that of first 6 terms is 125 : 152. Find the common ratio.

Let first term of G.P. be a and common ratio be r.

Given,

$\Rightarrow \dfrac{S$3}{S6} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{\dfrac{a(1 - r^3)}{1 - r}}{\dfrac{a(1 - r^6)}{1 - r}} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{(1 - r^3)}{(1 - r^6)} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{(1 - r^3)}{(1)^2 - (r^3)^2} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{(1 - r^3)}{(1 + r^3)(1 - r^3)} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{1}{(1 + r^3)} = \dfrac{125}{152} \\[1em] \Rightarrow 1 + r^3 = \dfrac{152}{125} \\[1em] \Rightarrow r^3 = \dfrac{152}{125} - 1 \\[1em] \Rightarrow r^3 = \dfrac{152 - 125}{125} \\[1em] \Rightarrow r^3 = \dfrac{27}{125} \\[1em] \Rightarrow r = \sqrt[3]{\dfrac{27}{125}} \\[1em] \Rightarrow r = \dfrac{3}{5}.

Hence, r = $\dfrac{3}{5}$.