Which term of the G.P.: (i) √3, 3, 3√3, … is 729 ? (ii) 1/3, 1/9, 1/27… is 1/19683 ?

Home » Which term of the G.P.: (i) √3, 3, 3√3, … is 729 ? (ii) 1/3, 1/9, 1/27… is 1/19683 ?

Which term of the G.P.:
(i) √3, 3, 3√3, … is 729 ?
(ii) 1/3, 1/9, 1/27… is 1/19683 ?

Which term of the G.P.:
(i) √3, 3, 3√3, … is 729 ?
(ii) 1/3, 1/9, 1/27… is 1/19683 ?

Solution:

(i) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is $729 ?$
Using the formula,
$\begin{array}{l} \mathrm{T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \\ \mathrm{a}=\sqrt{3} \\ \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(3 / \sqrt{3}) \\ =\sqrt{3} \\ \mathrm{~T}_{\mathrm{n}}=729 \\ \mathrm{n}=? \\ \mathrm{~T}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}-1} \\ 729=\sqrt{3}(\sqrt{3})^{\mathrm{n}-1} \\ 729=(\sqrt{3})^{\mathrm{n}} \\ 3^{6}=(\sqrt{3})^{\mathrm{n}} \\ (\sqrt{3})^{12}=(\sqrt{3})^{\mathrm{n}} \\ \mathrm{n}=12 \end{array}$
Therefore, $12^{\text {th }}$ term of the G.P is 729

(ii) $1 / 3,1 / 9,1 / 27 \ldots$ is $1 / 19683 ?$
Using the formula.
$\begin{array}{l} T_{n}=a r^{n-1} \\ a=1 / 3 \\ r=t_{2} / t_{1}=(1 / 9) /(1 / 3) \\ =1 / 9 \times 3 / 1 \end{array}$
$\begin{array}{l} =1 / 3 \\ T_{n}=1 / 19683 \\ n=? \\ T_{n}=a r^{n-1} \\ 1 / 19683=(1 / 3)(1 / 3)^{n-1} \\ 1 / 19683=(1 / 3)^{n} \\ (1 / 3)^{9}=(1 / 3)^{n} \\ n=9 \end{array}$
Therefore, $9^{\text {th }}$ term of the G.P is $1 / 19683$