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Show that the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term is $\frac{1}{r^n}$.
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Sum of the first $n$ terms of a G.P. $=a+a r+a r^2$ $+$................$+a r^{n-1}=\frac{a\left(1-r^n\right)}{1-r} \quad \ldots(i)$
$T_{n+1}=a r^n, T_{n+2}=a r^{n+1}, \ldots \ldots \ldots T_{2 n}=a r^{2 n-1}$
$\therefore \quad$ Sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ terms
$=T_{n+1}+T_{n+2}+T_{n+3}+\ldots \ldots \ldots . .+T_{2 n}$
$=a r^n+a r^{n+1}+a r^{n+2}+\ldots \ldots \ldots+a r^{2 n-1}$
$=\frac{a r^n\left(1-r^n\right)}{1-r} \quad \ldots(ii)$
No. of terms $=n$
Dividing (i) by (ii),
$=\frac{\frac{a\left(1-r^n\right)}{1-r}}{\frac{a r^n\left(1-r^n\right)}{1-r}}=\frac{1}{r^n}$
$T_{n+1}=a r^n, T_{n+2}=a r^{n+1}, \ldots \ldots \ldots T_{2 n}=a r^{2 n-1}$
$\therefore \quad$ Sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ terms
$=T_{n+1}+T_{n+2}+T_{n+3}+\ldots \ldots \ldots . .+T_{2 n}$
$=a r^n+a r^{n+1}+a r^{n+2}+\ldots \ldots \ldots+a r^{2 n-1}$
$=\frac{a r^n\left(1-r^n\right)}{1-r} \quad \ldots(ii)$
No. of terms $=n$
Dividing (i) by (ii),
$=\frac{\frac{a\left(1-r^n\right)}{1-r}}{\frac{a r^n\left(1-r^n\right)}{1-r}}=\frac{1}{r^n}$
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