The sum of the first three terms of a G.P. is | vedclass

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(C) Let the first term be $a$ and the common ratio be $r$.The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Give

Vedclass · Accepted Answer

$3$

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(C) Let the first term be $a$ and the common ratio be $r$.The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.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 $\frac{r^3 - 1}{r^6 - 1} = \frac{125}{152}$.Using the identity $r^6 - 1 = (r^3 - 1)(r^3 + 1)$,we get $\frac{r^3 - 1}{(r^3 - 1)(r^3 + 1)} = \frac{125}{152}$.$\frac{1}{r^3 + 1} = \frac{125}{152}$.$125(r^3 + 1) = 152$.$125r^3 + 125 = 152$.$125r^3 = 152 - 125 = 27$.$r^3 = \frac{27}{125} = (\frac{3}{5})^3$.Therefore,$r = 3/5$.

The sum of the first three terms of a $G.P.$ is to the sum of the first six terms as $125: 152$. Find the common ratio of the $G.P.$ (in $/5$)

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