Question

Find the sum of first $n$ terms and the sum of first $5$ terms of the geometric series $1+\frac{2}{3}+\frac{4}{9}+\cdots$

Answer 1

Given G.P : $1+\frac{2}{3}+\frac{4}{9}+\cdots$
Here, first term $a=1$
and common ratio, $r=\frac{\frac{2}{3}}{1}=\frac{2}{3}$

Step $1:$ Finding sum of $n$ terms of G.P
We know that, sum of $n$ terms of G.P.
$S_n=\frac{a(1-r^n)}{1-r}$
$\Rightarrow S_n=\frac{1(1-{\left(\frac{2}{3}\right)}^n)}{1-\frac{2}{3}}$
$\Rightarrow S_n=\frac{1-{\left(\frac{2}{3}\right)}^n}{\frac{1}{3}}$
$\therefore S_n=3[1-{\left(\frac{2}{3}\right)}^n]\cdots \left(i\right)$

Step $2:$ Finding sum of first $5$ terms.
Putting $n=5$ in equation $\left(i\right)$ we get
$S_5=3[1-{\left(\frac{2}{3}\right)}^5]$
$\Rightarrow S_5=3[1-\left(\frac{2^5}{3^5}\right)]$
$\Rightarrow S_5=3[1-\left(\frac{32}{243}\right)]$
$\Rightarrow S_5=3\left[\frac{243-32}{243}\right]$
$\therefore S_5=\frac{211}{81}$