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

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(C) Let the $G.P.$ have $2n$ terms with first term $a$ and common ratio $r$.The sum of all $2n$ terms is $S_{2n} = \frac{a(r^{2n} - 1)}{r - 1}$.The te

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$4$

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(C) Let the $G.P.$ have $2n$ terms with first term $a$ and common ratio $r$.The sum of all $2n$ terms is $S_{2n} = \frac{a(r^{2n} - 1)}{r - 1}$.The terms at odd places are $a, ar^2, ar^4, \dots, ar^{2n-2}$. This is a $G.P.$ with $n$ terms,first term $a$,and common ratio $r^2$.The sum of terms at odd places is $S_{odd} = \frac{a((r^2)^n - 1)}{r^2 - 1} = \frac{a(r^{2n} - 1)}{r^2 - 1}$.Given that $S_{2n} = 5 	imes S_{odd}$,we have:$\frac{a(r^{2n} - 1)}{r - 1} = 5 	imes \frac{a(r^{2n} - 1)}{r^2 - 1}$.Since $r 
eq 1$ and $r^{2n} 
eq 1$,we can cancel $\frac{a(r^{2n} - 1)}{r - 1}$ from both sides:$1 = \frac{5}{r + 1}$.$r + 1 = 5 \Rightarrow r = 4$.

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

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