**Question:**

In a GP, the ratio of the sum of the first three terms is to first six terms is 125: 152. Find the common ratio.

**Solution:**

Sum of a G.P. series is represented by the formula, $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$

hen r≠1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Sum of first 3 terms $={ }^{\mathrm{a}} \times \frac{\mathrm{r}^{3}-1}{\mathrm{r}-1}$

Sum of first 6 terms $=a \times \frac{r^{6}-1}{r-1}$

$\therefore \frac{a \times \frac{r^{3}-1}{r-1}}{a \times \frac{r^{6}-1}{r-1}}=\frac{125}{152}$

$\Rightarrow \frac{\left(r^{3}-1\right)}{\left(r^{6}-1\right)}=\frac{125}{152}$

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

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

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

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

$\Rightarrow\left(125 r^{3}-27\right)\left(r^{3}-1\right)=0$

Either $125 r^{3}-27=0$ or $r^{3}-1=0$

Either $125 r^{3}=27$ or $r^{3}=1$

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

Either $r=\frac{3}{5}$ or $r=1$

Since r ≠ 1 [ if r is 1, all the terms will be equal which destroys the purpose ]

$\therefore r=\frac{3}{5}$