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.

$$\begin{gathered} \therefore S_3=a\left(\frac{r^3-1}{r-1}\right)\mathrm{and}{S}_6=a\left(\frac{r^6-1}{r-1}\right) \\ \mathrm{Then},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question} \\ \frac{S_3}{S_6}=\frac{a\left(\frac{r^3-1}{r-1}\right)}{a\left(\frac{r^6-1}{r-1}\right)} \\ \Rightarrow \frac{125}{152}=\frac{r^3-1}{r^6-1} \\ \Rightarrow 125\left(r^6-1\right)=152\left(r^3-1\right) \\ \Rightarrow 125r^6-125=152r^3-152 \\ \Rightarrow 125r^6-152r{}^3+27=0 \\ \mathrm{Now},\mathrm{let}{r}^3=y \\ \therefore 125y^2-152y+27=0 \\ \mathrm{Now},\mathrm{applying}\mathrm{the}\mathrm{quadatic}\mathrm{formula} \\ y=\left\{\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right\} \\ \Rightarrow y=\left\{\frac{152\pm \sqrt{9604}}{250}\right\} \\ \Rightarrow y=\left\{\frac{152+\sqrt{9604}}{250}\right\}\mathrm{or}\left\{\frac{152-\sqrt{9604}}{250}\right\} \\ \Rightarrow y=1\mathrm{or}\frac{27}{125} \\ \therefore r^3=1\mathrm{or}\mathit{}{r}^3=\frac{27}{125} \\ \mathrm{But},\mathit{}r=1\mathrm{is}\mathrm{not}\mathrm{possible}. \\ \therefore \mathit{}r=\sqrt[3]{\frac{27}{125}}=\frac{3}{5} \end{gathered}$$