If the sum and product of four positive consecutive terms of a...

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 =${a}^{4}{r}^{6} = 1296$

$⇒ {a}^{2}{r}^{3} = 36$

$⇒ a = \dfrac{6}{{r}^{3 / 2}}$

Sum =$a + ar + a{r}^{2} + a{r}^{3} = 126$

$⇒ \dfrac{1}{{r}^{3 / 2}} + \dfrac{r}{{r}^{3 / 2}} + \dfrac{{r}^{2}}{{r}^{3 / 2}} + \dfrac{{r}^{3}}{{r}^{3 / 2}} = \dfrac{126}{6}$

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

Let ${r}^{1 / 2} + {r}^{- 1 / 2} = A$

$⇒ {r}^{- 3 / 2} + {r}^{3 / 2} = ( r\dfrac{1}{2} + r\dfrac{- 1}{2}{)}^{3} - 3 ( r\dfrac{1}{2} + r\dfrac{- 1}{2} )$

$= {A}^{3} - 3A$

Therefore,

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

$⇒ {A}^{3} - 2A = 21$

$⇒ A = 3$

$∴ \sqrt{r} + \dfrac{1}{\sqrt{r}} = 3$

$⇒ r + 1 = 3\sqrt{r}$

On squaring both sides:

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

$⇒ {r}^{2} - 7r + 1 = 0$

Here, sum of roots$= 7$