Question

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

Answer 1

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

\[\therefore S_3 = a\left( \frac{r^3 - 1}{r - 1} \right) \text { and  }S_6 = a\left( \frac{r^6 - 1}{r - 1} \right)\]

\[\text { Then, according to the question }\]

\[ \frac{S_3}{S_6} = \frac{a\left( \frac{r^3 - 1}{r - 1} \right)}{a \left( \frac{r^6 - 1}{r - 1} \right)} \]

\[ \Rightarrow \frac{125}{152} = \frac{r^3 - 1}{r^6 - 1}\]

\[ \Rightarrow 125 \left( r^6 - 1 \right) = 152 \left( r^3 - 1 \right)\]

\[ \Rightarrow 125 r^6 - 125 = 152 r^3 - 152\]

\[ \Rightarrow 125 r^6 - 152r {}^3 + 27 = 0\]

\[\text { Now,  let } r^3 = y \]

\[ \therefore 125 y^2 - 152y + 27 = 0\]

\[\text { Now, applying the quadatic formula }\]

\[y = \left\{ \frac{- b \pm \sqrt{b^2 - 4ac}}{2a} \right\} \]

\[ \Rightarrow y = \left\{ \frac{152 \pm \sqrt{9604}}{250} \right\}\]

\[ \Rightarrow y = \left\{ \frac{152 + \sqrt{9604}}{250} \right\} or \left\{ \frac{152 - \sqrt{9604}}{250} \right\}\]

\[ \Rightarrow y = 1 \text { or } \frac{27}{125}\]

\[ \therefore r^3 = 1\text {  or } r^3 = \frac{27}{125}\]

\[\text { But, r = 1 is not possible } . \]

\[ \therefore r = \sqrt[3]{\frac{27}{125}} = \frac{3}{5}\]