# Find:

Question:

Find:

(i) the ninth term of the G.P. 1, 4, 16, 64, ...

(ii) the 10 th term of the G.P. $-\frac{3}{4}, \frac{1}{2},-\frac{1}{3}, \frac{2}{9}, \ldots$

(iii) the 8th term of the G.P. 0.3, 0.06, 0.012, ...

(iv) the 12 th term of the G.P. $\frac{1}{a^{3} x^{3}}, a x, a^{5} x^{5}, \ldots$

(v) $n$th term of the G.P. $\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3 \sqrt{3}}, \ldots$

(vi) the 10th term of the G.P. $\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \ldots$

Solution:

(i)

Here,

First term, $a=1$

Common ratio, $r=\frac{a_{2}}{a_{1}}=\frac{4}{1}=4$

$\therefore 9$ th term $=a_{9}=a r^{(9-1)}=1(4)^{8}=4^{8}=65536$

Thus, the 9 th term of the given GP is 65536 .

(ii)

Here,

First term, $a=\frac{-3}{4}$

Common ratio, $r=\frac{a_{2}}{a_{1}}=\frac{\frac{1}{2}}{-\frac{3}{4}}=-\frac{2}{3}$

$\therefore 10$ th term $=a_{10}=a r^{(10-1)}=\left(\frac{-3}{4}\right)\left(\frac{-2}{3}\right)^{9}=\frac{1}{2}\left(\frac{2}{3}\right)^{8}$

Thus, the 10 th term of the given GP is $\frac{1}{2}\left(\frac{2}{3}\right)^{8}$.

(iii)

Here,

First term, $a=0.3$

Common ratio, $r=\frac{a_{2}}{a_{1}}=\frac{0.06}{0.3}=0.2$

$\therefore 8$ th term $=a_{8}=a r^{(8-1)}=0.3(0.2)^{7}$

Thus, the 8 th term of the given GP is $0.3(0.2)^{7}$.

(iv)

Here,

First term, $a=\frac{1}{a^{3} x^{3}}$

Common ratio, $r=\frac{a_{2}}{a_{1}}=\frac{a x}{\frac{1}{a^{3} x^{3}}}=a^{4} x^{4}$

$\therefore 12$ th term $=a_{12}=a r^{(12-1)}=\frac{1}{a^{3} x^{3}}\left(a^{4} x^{4}\right)^{11}=a^{41} x^{41}$

Thus, the 12 th term of the given GP is $a^{41} x^{41}$.

(v)

Here,

First term, $a=\sqrt{3}$

Common ratio, $r=\frac{a_{2}}{a_{1}}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{3}$

$\therefore n$th term $=a_{n}=a r^{(n-1)}=\sqrt{3}\left(\frac{1}{3}\right)^{n-1}$

Thus, the $n$th term of the given GP is $\sqrt{3}\left(\frac{1}{3}\right)^{n-1}$.

(vi)

Here,

First term, $a=\sqrt{2}$

Common ratio, $r=\frac{a_{2}}{a_{1}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$

$\therefore 10$ th term $=a_{10}=a r^{(10-1)}=\sqrt{2}\left(\frac{1}{2}\right)^{9}=\frac{1}{\sqrt{2}} \times \frac{1}{2^{8}}$

Thus, the 10 th term of the given GP is $\frac{1}{\sqrt{2}} \times \frac{1}{2^{8}}$.