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

• A. $\frac{9}{2}$
• B. 3
• C. 7
• D. 14

Answer 1

Answer: (C)

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