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

Question

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

Accepted Answer

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

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

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

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

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

$\frac{1}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^{2}}{\mathrm{r}^{3 / 2}}+\frac{\mathrm{r}^{3}}{\mathrm{r}^{3 / 2}}=\frac{126}{6}=21$

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

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

$\mathrm{r}^{-3 / 2}+\mathrm{r}^{3 / 2}+3 \mathrm{~A}=\mathrm{A}^{3}$

$\mathrm{A}^{3}-3 \mathrm{~A}+\mathrm{A}=21$

$\mathrm{A}^{3}-2 \mathrm{~A}=21$

$A=3$

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

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

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

$r^{2}-7 r+1=0$

$\Rightarrow r_{1}+r_{2}=7$

Solutions · 1