If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then…

Let us consider four positive consecutive terms of a G.P. as follow:

$a,ar,a{r}^2,a{r}^3\left(a,r>0\right)$

Product $=$$a^4{r}^6=1296$

$\Rightarrow a^2{r}^3=36$

$\Rightarrow a=\frac{6}{r^{3/2}}$

Sum $=$$a+ar+a{r}^2+a{r}^3=126$

$\Rightarrow \frac{1}{r^{3/2}}+\frac{r}{r^{3/2}}+\frac{r^2}{r^{3/2}}+\frac{r^3}{r^{3/2}}=\frac{126}{6}$

$\Rightarrow \left(r^{-3/2}+r^{3/2}\right)+\left(r^{1/2}+r^{-1/2}\right)=21$

Let $r^{1/2}+r^{-1/2}=A$

$\Rightarrow r^{-3/2}+r^{3/2}=(r^{\frac{1}{2}}+r^{\frac{-1}{2}}{)}^3-3(r^{\frac{1}{2}}+r^{\frac{-1}{2}})$

$=A^3-3A$

Therefore,

$A^3-3A+A=21$

$\Rightarrow A^3-2A=21$

$\Rightarrow A=3$

$\therefore \sqrt{r}+\frac{1}{\sqrt{r}}=3$

$\Rightarrow r+1=3\sqrt{r}$

On squaring both sides:

$r^2+2r+1=9r$

$\Rightarrow r^2-7r+1=0$

Here, sum of roots$=7$