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Questions
The ratio of the sum of the first three terms to that of the first 6 terms of a G.P. is 125 : 152. Find the common ratio.
The ratio of the sum of the first three terms to the sum of the first six terms of a G.P. is 125 : 152. Find the common ratio of G.P.
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Solution
Let a be the first term and r be the common ratio of the G.P.
\[\therefore S_3 = a\left( \frac{r^3 - 1}{r - 1} \right) \text { and }S_6 = a\left( \frac{r^6 - 1}{r - 1} \right)\]
Then, according to the question
\[ \frac{S_3}{S_6} = \frac{a\left( \frac{r^3 - 1}{r - 1} \right)}{a \left( \frac{r^6 - 1}{r - 1} \right)} \]
\[ \Rightarrow \frac{125}{152} = \frac{r^3 - 1}{r^6 - 1}\]
\[ \Rightarrow 125 \left( r^6 - 1 \right) = 152 \left( r^3 - 1 \right)\]
\[ \Rightarrow 125 r^6 - 125 = 152 r^3 - 152\]
\[ \Rightarrow 125 r^6 - 152r {}^3 + 27 = 0\]
\[\text { Now, let } r^3 = y \]
\[ \therefore 125 y^2 - 152y + 27 = 0\]
Now, applying the quadratic formula
\[y = \left\{ \frac{- b \pm \sqrt{b^2 - 4ac}}{2a} \right\} \]
\[ \Rightarrow y = \left\{ \frac{152 \pm \sqrt{9604}}{250} \right\}\]
\[ \Rightarrow y = \left\{ \frac{152 + \sqrt{9604}}{250} \right\} or \left\{ \frac{152 - \sqrt{9604}}{250} \right\}\]
\[ \Rightarrow y = 1 \text { or } \frac{27}{125}\]
\[ \therefore r^3 = 1\text { or } r^3 = \frac{27}{125}\]
But, r = 1 is not possible
\[ \therefore r = \sqrt[3]{\frac{27}{125}} = \frac{3}{5}\]
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