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

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$4$

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(C) Let the $G.P.$ have $2n$ terms,with the first term $a$ and common ratio $r$.The sum of all $2n$ terms is $S_{2n} = a\frac{(r^{2n} - 1)}{(r - 1)}$.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 $S_{odd} = a\frac{((r^2)^n - 1)}{(r^2 - 1)} = a\frac{(r^{2n} - 1)}{(r^2 - 1)}$.According to the problem,$S_{2n} = 5 	imes S_{odd}$.Substituting the formulas: $a\frac{(r^{2n} - 1)}{(r - 1)} = 5 	imes a\frac{(r^{2n} - 1)}{(r^2 - 1)}$.Since $r 
eq 1$ and $r^{2n} - 1 
eq 0$,we can divide both sides by $a\frac{(r^{2n} - 1)}{(r - 1)}$:$1 = \frac{5}{r + 1}$.$r + 1 = 5$.$r = 4$.

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

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