Question

Which term of the following sequence:
$\left(i\right)2,2\sqrt{2},4,...$ is $128?$
$\left(ii\right)\sqrt{3},3,3\sqrt{3},...$ is $729?$
$\left(iii\right)\frac{1}{3},\frac{1}{9},\frac{1}{27},...$ is $\frac{1}{19683}?$

Answer 1

$\left(i\right)$ Given : $2,2\sqrt{2},4,...$
Here, first term $a=2$
common ration $r=\frac{2\sqrt{2}}{2}=\sqrt{2}$
and $n^{th}$ term $=128$
We know, the $n^{th}$ term of G.P.is
$a_n=a{r}^{n-1}$
$\Rightarrow 128=2\times (\sqrt{2}{)}^{n-1}$
$\Rightarrow 64=\left(\right(\sqrt{2}{)}^{n-1}$
$\Rightarrow (\sqrt{2}{)}^{12}=(\sqrt{2}{)}^{n-1}$
$\Rightarrow 12=n-1$
$\therefore n=13$

$\left(ii\right)$ Given: $\sqrt{3},3,3\sqrt{3},...$
Here, first term $a=\sqrt{3}$,
common ratio $r=\frac{3}{\sqrt{3}}=\sqrt{3}$
and $n^{th}$ term $=729$
We know, the $n^{th}$ term of G.P. is
$a_n=a{r}^{n-1}$
$\Rightarrow 729=\sqrt{3}\times (\sqrt{3}{)}^{n-1}$
$\Rightarrow 729=(\sqrt{3}{)}^n$
$\Rightarrow (\sqrt{3}{)}^{12}=(\sqrt{3}{)}^n$
By comparing powers, we get
$n=12$

$\left(iii\right)$ Given: $\frac{1}{3},\frac{1}{9},\frac{1}{27}...$
Here, first term $a=\frac{1}{3}$,
common ratio $r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{9}\times \frac{3}{1}=\frac{1}{3}$
and $n^{th}$ term $=\frac{1}{19683}$

We know, the $n^{th}$ term of G.P. is
$a_n=a{r}^{n-1}$
$\Rightarrow \frac{1}{19683}=\frac{1}{3}\times {\left(\frac{1}{3}\right)}^{n-1}$
$\Rightarrow \frac{1}{19683}=\frac{1}{3^n}$
$\Rightarrow {\left(\frac{1}{3}\right)}^9={\left(\frac{1}{3}\right)}^n$
By comparing powers, we get
$\therefore n=9$