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, a r, 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+a r+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}+3 A=A^3$
$ A ^3-3 A + A =21 $
$ A ^3-2 A =21 $
$ A =3 $
$ \sqrt{r}+\frac{1}{\sqrt{r}}=3 $
$ r +1=3 \sqrt{r} $
$ r ^2+2 r +1=9 r $
$ r ^2-7 r +1=0$