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# Show that the ratio of the first $n$ terms of a G.P. to the sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ term is $1:r^n$

Toolbox:
• Sum of first $n$ terms of a G.P$=a.\large\frac{1-r^n}{1-r}$
Step 1
Given that the G.P. has $2n$ terms.
Let the first $2n$ terms of the G.P be
$(a+ar+ar^2+......ar^{n-1})+(ar^n+ar^{n+1}+....ar^{2n-1})$
Let the sum of first $n$ terms be $S_n$,
Sum of first $2n$ terms be $S_{2n}$ and
Sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ term be $S$
We know that the sum of first $n$ terms of a G.P$=S_n=a.\large\frac{1-r^n}{1-r}$
$\Rightarrow\:$The sum of first $2n$ terms of the G.P$=S_{2n}=a.\large\frac{1-r^{2n}}{1-r}$
$\therefore\:S_{2n}=(a+ar+ar^2+......ar^{n-1})+(ar^n+ar^{n+1}+....ar^{2n-1})$
$\Rightarrow\:S_{2n}=S_n+S$
$\Rightarrow\:a.\large\frac{1-a^{2n}}{1-r}$$=a.\large\frac{1-r^n}{1-r}$$+S$
$\Rightarrow\:$Sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ term $=S=$
$(a.r^{n}+a.r^{n+1}+........a.r^{2n-1})=S=a.\large\frac{1-a^{2n}}{1-r}$$-a.\large\frac{1-r^n}{1-r} Taking a\:\;and\:\:(1-r) common we get \Rightarrow\:S=\large\frac{a}{1-r}$$.(1-r^{2n}-1+r^n)$
$\Rightarrow\:S=\large\frac{a}{1-r}$$.(r^n-r^{2n}) Taking r^n common we get \Rightarrow\:S=\large\frac{ar^n}{1-r}$$.(1-r^{n})$
Step 2
The ratio of the first $n$ terms of a G.P. to the sum of the terms from $(n+1)^{th}$ to $(2n)^{th}$ term =
$S_n\: : \:S=a.\large\frac{1-r^n}{1-r}\: :\: \frac{a.r^n}{1-r}.$$(1-r^n)$
Cancelling the terms $a,\:\:(1-r^n)\:\:and\:\:(1-r)$ on both the sides we get
$\Rightarrow\:S_n\: :\:S=1\: :\:r^n$
Hence Proved.
edited Mar 3, 2014