Question

Find :

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

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

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

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

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

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

Answer 1

(i) Here,

First term, a = 1

Common ratio, $r=\frac{a_2}{a_1}=\frac{4}{1}=4$

$\therefore$ 9th term $=a_9=a{r}^{(9-1)}=1(4{)}^8=4^8=65536$

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

(ii) Here,

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

Common ratio, $r=\frac{a_2}{a_1}=\frac{\frac{1}{2}}{-\frac{3}{4}}=-\frac{2}{3}$

10th term $=a_{1}0=a{r}^{(10-1)}=\left(\frac{-3}{4}\right){\left(\frac{-2}{3}\right)}^9=\frac{1}{2}{\left(\frac{2}{3}\right)}^8$

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

(iii) Here,

First term, a = 0.3

Common ratio, $r=\frac{a_2}{a_1}=\frac{0.06}{0.3}=0.2$

$\therefore$ 8th term $=a_8=a{r}^{(8-1)}=0.3(0.2{)}^7$

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

(iv) Here,

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

Common ratio, $r=\frac{a_2}{a_1}=\frac{\frac{ax}{1}}{a^3{x}^3}=a^4{x}^4$

$\therefore$ 12th term $=a_{1}2=a{r}^{(12-1)}=\frac{1}{a^3{x}^3}(a^4{x}^4{)}^{11}=a^{41}{x}^{41}$

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

(v) Here,

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

Common ratio, $r=\frac{a_2}{a_1}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{3}$

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

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

(iv) Here,

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

Common ratio, $r=\frac{a_2}{a_1}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$

$\therefore$ 10th term $=a_{1}0=a{r}^{(10-1)}=\sqrt{2}{\left(\frac{1}{2}\right)}^9$

$=\frac{1}{\sqrt{2}}\times \frac{1}{2^8}$

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