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 $\frac{1}{19683}$ ?
(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 $\frac{1}{19683}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.