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

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8. Sequences and Series

hard

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 $.........$.

$7$

$\frac{9}{2}$

$3$

$14$

(JEE MAIN-2023)

Solution

$a, a r, a r^2, a r^3(a, r > 0)$

$a^4 r^6=1296$

$a^2 r^3=36$

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

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

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

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

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

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

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

$A ^3-2 A =21$

$A =3$

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

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

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

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

Standard 11

Mathematics

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Solution

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