If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is

$\mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}, \mathrm{ar}^{3}(\mathrm{a}, \mathrm{r}>0)$ <br/><br/>$a^{4} r^{6}=1296$ <br/><br/>$a^{2} r^{3}=36$ <br/><br/>$a=\frac{6}{r^{3 / 2}}$ <br/><br/>$a+a r+a r^{2}+a r^{3}=126$ <br/><br/>$\frac{1}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^{2}}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^{3}}{\mathrm{r}^{3 / 2}}=\frac{126}{6}=21$ <br/><br/>$\left(r^{-3 / 2}+r^{3 / 2}\right)+\left(r^{1 / 2}+r^{-1 / 2}\right)=21$ <br/><br/>$\mathrm{r}^{1 / 2}+\mathrm{r}^{-1 / 2}=\mathrm{A}$ <br/><br/>$\mathrm{r}^{-3 / 2}+\mathrm{r}^{3 / 2}+3 \mathrm{~A}=\mathrm{A}^{3}$ <br/><br/>$\mathrm{A}^{3}-3 \mathrm{~A}+\mathrm{A}=21$ <br/><br/>$\mathrm{A}^{3}-2 \mathrm{~A}=21$ <br/><br/>$A=3$ <br/><br/>$\sqrt{\mathrm{r}}+\frac{1}{\sqrt{\mathrm{r}}}=3$ <br/><br/>$\mathrm{r}+1=3 \sqrt{\mathrm{r}}$ <br/><br/>$r^{2}+2 r+1=9 r$ <br/><br/>$r^{2}-7 r+1=0$ <br/><br/>$\Rightarrow r_{1}+r_{2}=7$