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 [2023]

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Solution

Q.
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 [2023]

1 $\frac{9}{2}$

2 3

3 7

4 14

Ans.
(3)

$Let the terms be \frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}$

$According to question,$ $\frac{a}{r^{3}} \cdot \frac{a}{r} \cdot a r \cdot a r^{3} = 1296$ $\Rightarrow a = 6$

Now, $\frac{a}{r^{3}} + \frac{a}{r} + a r + a r^{3} = 126$ $\Rightarrow \frac{1}{r^{3}} + \frac{1}{r} + r + r^{3} = 21$

$\Rightarrow {(r + \frac{1}{r})}^{3} - 2 (r + \frac{1}{r}) = 21$

$Let r + \frac{1}{r} = t \Rightarrow t^{3} - 2 t - 21 = 0$

$\Rightarrow t = 3 \Rightarrow r + \frac{1}{r} = 3 \Rightarrow r^{2} - 3 r + 1 = 0$

$Sum of roots = r_{1}^{2} + r_{2}^{2} = 7$

$Hence, the positive sum is 7 .$

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