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(A) Given the sum of the first $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.We are given $\frac{S_3}{S_6} = \frac{125}{152}$.Substituting

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$\frac{3}{5}$

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(A) Given the sum of the first $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.We are given $\frac{S_3}{S_6} = \frac{125}{152}$.Substituting the formula: $\frac{a(r^3 - 1)/(r - 1)}{a(r^6 - 1)/(r - 1)} = \frac{125}{152}$.This simplifies to $\frac{r^3 - 1}{r^6 - 1} = \frac{125}{152}$.Since $r^6 - 1 = (r^3 - 1)(r^3 + 1)$,we have $\frac{1}{r^3 + 1} = \frac{125}{152}$.$152 = 125(r^3 + 1) \Rightarrow 152 = 125r^3 + 125$.$125r^3 = 152 - 125 = 27$.$r^3 = \frac{27}{125} = (\frac{3}{5})^3$.Therefore,$r = \frac{3}{5}$.

If the ratio of the sum of the first three terms and the sum of the first six terms of a $G.P.$ is $125 : 152$,then the common ratio $r$ is:

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