Show that the ratio of the sum of first $n$ terms of a $G.P.$ to the sum of terms from
$(n+1)^{ th }$ to $(2 n)^{ th }$ term is $\frac{1}{r^{n}}$
Show that the ratio of the sum of first $n$ terms of a $G.P.$ to the sum of terms from
$(n+1)^{ th }$ to $(2 n)^{ th }$ term is $\frac{1}{r^{n}}$
Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
Sum of first $n$ terms $=\frac{a\left(1-r^{n}\right)}{(1-r)}$
Since there are $n$ terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term,
Sum of terms from $(n+1)^{t h}$ to $(2 n)^{th}$ term
$S_{n}=\frac{a_{n+1}\left(1-r^{n}\right)}{1-r}$
$a^{n+1}=a r^{n+1-1}=a r^{n}$
Thus, required ratio $=\frac{a\left(1-r^{n}\right)}{(1-r)} \times \frac{(1-r)}{a r^{n}\left(1-r^{n}\right)}=\frac{1}{r^{n}}$
Thus, the ratio of the sum of first $n$ terms of a $G.P.$ to the sum of terms from term is $\frac{1}{r^{n}}$
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