If S_n stands for the sum of n terms of a GP | vedclass

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(B) The sum of $n$ terms of a $GP$ is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.Similarly,the sum of $2n$ terms is $S_{2n} = \frac{a(1 - r^{2n})}{1 -

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$\frac{1}{r^n + 1}$

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(B) The sum of $n$ terms of a $GP$ is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.Similarly,the sum of $2n$ terms is $S_{2n} = \frac{a(1 - r^{2n})}{1 - r}$.Taking the ratio,we have $\frac{S_n}{S_{2n}} = \frac{a(1 - r^n)}{1 - r} 	imes \frac{1 - r}{a(1 - r^{2n})}$.Using the identity $1 - r^{2n} = (1 - r^n)(1 + r^n)$,we get $\frac{S_n}{S_{2n}} = \frac{1 - r^n}{(1 - r^n)(1 + r^n)}$.Simplifying this,we obtain $\frac{S_n}{S_{2n}} = \frac{1}{1 + r^n}$.

If $S_n$ stands for the sum of $n$ terms of a $GP$ with $a$ as the first term and $r$ as the common ratio,then $S_n : S_{2n}$ is

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