A G.P. consists of an even number of terms. I | vedclass

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(B) Let the $G.P.$ be $a, ar, ar^2, ar^3, \dots, ar^{2n-1}$.Number of terms $= 2n$.Sum of all terms $S_{2n} = \frac{a(r^{2n}-1)}{r-1}$.Sum of terms at

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$4$

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(B) Let the $G.P.$ be $a, ar, ar^2, ar^3, \dots, ar^{2n-1}$.Number of terms $= 2n$.Sum of all terms $S_{2n} = \frac{a(r^{2n}-1)}{r-1}$.Sum of terms at odd places $S_{odd} = a + ar^2 + \dots + ar^{2n-2} = \frac{a((r^2)^n - 1)}{r^2 - 1} = \frac{a(r^{2n}-1)}{r^2-1}$.Given that $S_{2n} = 5 	imes S_{odd}$.$\frac{a(r^{2n}-1)}{r-1} = 5 	imes \frac{a(r^{2n}-1)}{r^2-1}$.Since $r 
eq 1$ and $r^{2n} 
eq 1$,we can cancel $\frac{a(r^{2n}-1)}{r-1}$ from both sides.$1 = \frac{5}{r+1}$.$r+1 = 5$.$r = 4$.

$A$ $G.P.$ consists of an even number of terms. If the sum of all the terms is $5$ times the sum of terms occupying odd places,then find its common ratio.

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