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

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Here}, \\ \mathrm{First}\mathrm{term},\mathit{}a=1 \\ \mathrm{Common}\mathrm{ratio},r=\frac{a_2}{a_1}=\frac{4}{1}=4 \\ \therefore 9\mathrm{th}\mathrm{term}=a_9=a{r}^{(9-1)}=1(4{)}^8=4^8=65536 \\ \mathrm{Thus},\mathrm{the}9\mathrm{th}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{GP}\mathrm{is}65536. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Here}, \\ \mathrm{First}\mathrm{term},\mathit{}a=\frac{-3}{4} \\ \mathrm{Common}\mathrm{ratio},r=\frac{a_2}{a_1}=\frac{{\displaystyle \frac{1}{2}}}{-{\displaystyle \frac{3}{4}}}=-\frac{2}{3} \\ \therefore 10\mathrm{th}\mathrm{term}=a_{10}=a{r}^{(10-1)}=\left(\frac{-3}{4}\right){\left(\frac{-2}{3}\right)}^9=\frac{1}{2}{\left(\frac{2}{3}\right)}^8 \\ \mathrm{Thus},\mathrm{the}10\mathrm{th}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{GP}\mathrm{is}\frac{1}{2}{\left(\frac{2}{3}\right)}^8. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{Here}, \\ \mathrm{First}\mathrm{term},\mathit{}a=0.3 \\ \mathrm{Common}\mathrm{ratio},r=\frac{a_2}{a_1}=\frac{0.06}{0.3}=0.2 \\ \therefore 8\mathrm{th}\mathrm{term}=a_8=a{r}^{(8-1)}=0.3(0.2{)}^7 \\ \mathrm{Thus},\mathrm{the}8\mathrm{th}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{GP}\mathrm{is}0.3(0.2{)}^7. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \\ \mathrm{Here}, \\ \mathrm{First}\mathrm{term},\mathit{}a=\frac{1}{a^3{x}^3} \\ \mathrm{Common}\mathrm{ratio},r=\frac{a_2}{a_1}=\frac{ax}{{\displaystyle \frac{1}{a^3{x}^3}}}=a^4{x}^4 \\ \therefore 12\mathrm{th}\mathrm{term}=a_{12}=a{r}^{(12-1)}=\frac{1}{a^3{x}^3}(a^4{x}^4{)}^{11}=a^{41}{x}^{41} \\ \mathrm{Thus},\mathrm{the}12\mathrm{th}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{GP}\mathrm{is}{a}^{41}{x}^{41}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{v}\right) \\ \mathrm{Here}, \\ \mathrm{First}\mathrm{term},\mathit{}a=\sqrt{3} \\ \mathrm{Common}\mathrm{ratio},r=\frac{a_2}{a_1}=\frac{{\displaystyle \frac{1}{\sqrt{3}}}}{\sqrt{3}}=\frac{1}{3} \\ \therefore n\mathrm{th}\mathrm{term}=a_n=a{r}^{(n-1)}=\sqrt{3}{\left(\frac{1}{3}\right)}^{n-1} \\ \mathrm{Thus},\mathrm{the}n\mathrm{th}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{GP}\mathrm{is}\sqrt{3}{\left(\frac{1}{3}\right)}^{n-1}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right) \\ \mathrm{Here}, \\ \mathrm{First}\mathrm{term},\mathit{}a=\sqrt{2} \\ \mathrm{Common}\mathrm{ratio},r=\frac{a_2}{a_1}=\frac{{\displaystyle \frac{1}{\sqrt{2}}}}{\sqrt{2}}=\frac{1}{2} \\ \therefore 10\mathrm{th}\mathrm{term}=a_{10}=a{r}^{(10-1)}=\sqrt{2}{\left(\frac{1}{2}\right)}^9=\frac{1}{\sqrt{2}}\times \frac{1}{2^8} \\ \mathrm{Thus},\mathrm{the}10\mathrm{th}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{GP}\mathrm{is}\frac{1}{\sqrt{2}}\times \frac{1}{2^8}. \end{gathered}$$