Question

Which term of the following sequences :

(a) $2,2\sqrt{2},4.....\text{is}128?$

(b) $\sqrt{3},3,3\sqrt{3},.....\text{is}729?$

c) $\frac{1}{3},\frac{1}{9},\frac{1}{27},.....\text{is}\frac{1}{19683}?$

Answer 1

(a) Here a = 2, and r = $\frac{2\sqrt{2}}{2}=\sqrt{2}.$

Let $n^{th}$ term of given sequence be 128 is, $a_n=128.$

We know that $a_n=a{r}^{n-1}$

$\therefore 128=2\times (\sqrt{2}{)}^{n-1}\Rightarrow 64=(\sqrt{2}{)}^{n-1}$

$\Rightarrow (\sqrt{2}{)}^{12}=(\sqrt{2}{)}^{n-1}\Rightarrow n-1=12$

$\Rightarrow$ n = 13

Thus, ${13}^{th}$ term of the given sequences is 128.

(b) Here $a=\sqrt{3}andr=\frac{3}{\sqrt{3}}=\sqrt{3}.$

Let $n^{th}$ term of given sequence be 729 i.e

$a_n$ = 729.

We know that $a_n=a{r}^{n-1}$

$\therefore 729=\sqrt{3}\times (\sqrt{3}{)}^{n-1}\Rightarrow (\sqrt{3}{)}^{12}=(\sqrt{3}{)}^n$

$\Rightarrow n=12$

Thus, the ${12}^{th}$ term of the given sequence is 729.

(c) Here, a = $\frac{1}{3}andr=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}$

Let $n^{th}$ term of given sequence be $\frac{1}{19683}.$

i.e. $a_n=\frac{1}{19683}$

We know that $a_n=a{r}^{n-1}$

$\therefore \frac{1}{19683}=\frac{1}{3}\times {\left(\frac{1}{3}\right)}^{n-1}\Rightarrow {\left(\frac{1}{3}\right)}^9={\left(\frac{1}{3}\right)}^n$

$\Rightarrow n=9$

Thus, the $9^{th}$ term of the given sequence is $\frac{1}{19683}.$