Which term of the G.P. :
Question:

Which term of the G.P. :

(i) $\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \frac{1}{4 \sqrt{2}}, \ldots$ is $\frac{1}{512 \sqrt{2}}$ ?

(ii) $2,2 \sqrt{2}, 4, \ldots$ is 128 ?

(iii) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is 729 ?

(iv) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27} \ldots$ is $\frac{1}{19683}$ ?

Solution:

(i)

Here, first term, $a=\sqrt{2}$

and common ratio, $r=\frac{1}{2}$

Let the $n^{\text {th }}$ term be $\frac{1}{512 \sqrt{2}}$.

$\therefore a_{n}=\frac{1}{512 \sqrt{2}}$

$\Rightarrow a r^{n-1}=\frac{1}{512 \sqrt{2}}$

$\Rightarrow(\sqrt{2})\left(\frac{1}{2}\right)^{n-1}=\frac{1}{512 \sqrt{2}}$

$\Rightarrow\left(\frac{1}{2}\right)^{n-1}=\frac{1}{1024}$

$\Rightarrow\left(\frac{1}{2}\right)^{n-1}=\left(\frac{1}{2}\right)^{10}$

$\Rightarrow n-1=10$

$\Rightarrow n=11$

Thus, the $11^{\text {th }}$ term of the given G.P. is $\frac{1}{512 \sqrt{2}}$.

(ii)

Here, first term, $a=2$

and common ratio, $r=\sqrt{2}$

Let the $n^{\text {th }}$ term be 128 .

$\therefore a_{n=128}$

$\Rightarrow a r^{n-1}=128$

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

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

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

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

$\Rightarrow n-1=12$

$\Rightarrow n=13$

Thus, the $13^{\text {th }}$ term of the given G. P. is 128 .

(iii)

Here, first term, $a=\sqrt{3}$

and common ratio, $r=\sqrt{3}$

Let the $n^{\text {th }}$ term be 729 .

$\therefore a_{n}=729$

$\Rightarrow a r^{n-1}=729$

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

$\Rightarrow(\sqrt{3})^{n-1}=\frac{(\sqrt{3})^{12}}{\sqrt{3}}=(\sqrt{3})^{11}$

$\Rightarrow n-1=11$

$\Rightarrow n=12$

Thus, the $12^{\text {th }}$ term of the given G. P. is 729 .

(iv)

Here, first term, $a=\frac{1}{3}$

and common ratio, $r=\frac{1}{3}$

Let the $n^{\text {th }}$ term be $\frac{1}{19683}$.

$\therefore a_{n}=\frac{1}{19683}$

$\Rightarrow a r^{n-1}=\frac{1}{19683}$

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

$\Rightarrow\left(\frac{1}{3}\right)^{n-1}=\frac{3}{(3)^{9}}=\left(\frac{1}{3}\right)^{8}$

$\Rightarrow n-1=8$

$\Rightarrow n=9$

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