Which term of the following sequences:
Question:

Which term of the following sequences:

(a) $2,2 \sqrt{2}, 4, \ldots$ is 128 ?

(b) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is 729 ?

(c) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$ ?

Solution:

(a) The given sequence is $2,2 \sqrt{2}, 4, \ldots$

Here, $a=2$ and $r=\frac{2 \sqrt{2}}{2}=\sqrt{2}$

Let the $n^{\text {th }}$ term of the given sequence be 128 .

$a_{n}=a r^{n-1}$

$\Rightarrow(2)(\sqrt{2})^{n-1}=128$

$\Rightarrow(2)(2)^{\frac{n-1}{2}}=(2)^{7}$

$\Rightarrow(2)^{\frac{n-1}{2}+1}=(2)^{7}$

$\therefore \frac{n-1}{2}+1=7$

$\Rightarrow \frac{n-1}{2}=6$

$\Rightarrow n-1=12$

$\Rightarrow n=13$

Thus, the $13^{\text {th }}$ term of the given sequence is 128 .

(b) The given sequence is $\sqrt{3}, 3,3 \sqrt{3}, \ldots$

Here, $a=\sqrt{3}$ and $r=\frac{3}{\sqrt{3}}=\sqrt{3}$

Let the $n^{\text {th }}$ term of the given sequence be 729 .

$a_{n}=a r^{n-1}$

$\therefore a r^{n-1}=729$

$\Rightarrow(\sqrt{3})(\sqrt{3})^{n-1}=729$

$\Rightarrow(3)^{\frac{1}{2}}(3)^{\frac{n-1}{2}}=(3)^{6}$

$\Rightarrow(3)^{\frac{1}{2}+\frac{n-1}{2}}=(3)^{6}$

$\therefore \frac{1}{2}+\frac{n-1}{2}=6$

$\Rightarrow \frac{1+n-1}{2}=6$

$\Rightarrow n=12$

Thus, the $12^{\text {th }}$ term of the given sequence is 729 .

(c) The given sequence is $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$

Here, $a=\frac{1}{3}$ and $r=\frac{1}{9} \div \frac{1}{3}=\frac{1}{3}$

Let the $n^{\text {th }}$ term of the given sequence be $\frac{1}{19683}$.

$a_{n}=a r^{n-1}$

$\therefore a r^{n-1}=\frac{1}{19683}$

$\Rightarrow\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{n-1}=\frac{1}{19683}$

$\Rightarrow\left(\frac{1}{3}\right)^{n}=\left(\frac{1}{3}\right)^{9}$

$\Rightarrow n=9$

Thus, the $9^{\text {th }}$ term of the given sequence is $\frac{1}{19683}$

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