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.

$\therefore S_3=a\left(\frac{r^3-1}{r-1}\right)and{S}_6=a\left(\frac{r^6-1}{r-1}\right)$

Then, according to the question,

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

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

$\Rightarrow 125(r^6-1)=152(r^3-1)$

$\Rightarrow 125r^6-125=152r^3-152$

$\Rightarrow 125r^6-152r^3+27=0$

Now, let $r^3=y$

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

Now, applying the quadratic formula

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

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

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

$\Rightarrow y=1or\frac{27}{125}$

$\therefore r^3=1or{r}^3=\frac{27}{125}$

But, r = 1 is not possible.

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