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  • 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 isOpti

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

Option: 1

7


Option: 2

3


Option: 3

\frac{9}{2}


Option: 4

14


Answers (1)

best_answer

Four term of G.P.   \frac{a}{r^{3}}\, ,\, \frac{a}{r},ar\, ,ar^{3} 

\begin{aligned} & \frac{a}{r^3}+\frac{a}{r}+a r+a r^3=126 \\ & \frac{a}{r^3} \cdot \frac{a}{r} \cdot a r \cdot a r^3=1296 \\ & a^4=1296 \\ & a=6 \\ & \frac{6}{r^3}+\frac{6}{r}+6 r+6 r^3=126 \end{aligned}

\begin{aligned} & \left(r+\frac{1}{r}\right)+r^3+\frac{1}{r^3}=21 \\ & \left(r+\frac{1}{r}\right)+\left(r+\frac{1}{r}\right)^3-3\left(r+\frac{1}{r}\right)=21 \\ & \text { Let } r+\frac{1}{r}=t \\ & t^3-2 t=21 \\ & \Rightarrow t=3 \end{aligned}

\begin{aligned} & r+\frac{1}{r}=3 \\ & r^2-3 r+1=0 \\ & r=\frac{3 \pm \sqrt{9-4}}{2} \\ & r=\frac{3 \pm \sqrt{5}}{2} \end{aligned}

Sum of common ratio  \frac{9}{4}+\frac{5}{4}+\frac{3 \sqrt{5}}{2}+\frac{9}{4}+\frac{5}{4}-\frac{3 \sqrt{5}}{2} 

= 7

Posted by

Ritika Harsh

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