Which term of the following sequences :

(a) $2,2\sqrt{2},4,...is128?$

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

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

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

Clearly given sequence is in G.P

With $a=2$ and $r=\frac{2\sqrt{2}}{2}=\sqrt{2}$

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

$$\begin{gathered} a_n={ar}^{n-1}\Rightarrow (2){(\sqrt{2})}^{n-1}=128 \\ \Rightarrow (2){(2)}^{\frac{n-1}{2}}={(2)}^7 \\ \Rightarrow (2{)}^{\frac{n-1}{2}+1}={(2)}^7 \\ \Rightarrow \frac{n-1}{2}=6\Rightarrow n-1=12\Rightarrow n=13 \end{gathered}$$

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

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

Clearly this sequence is in G.P

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

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

$$\begin{gathered} a_n={ar}^{n-1}\Rightarrow {ar}^{n-1}=729 \\ \Rightarrow (\sqrt{3}){(\sqrt{3})}^{n-1}=729 \\ \Rightarrow {(3)}^{\frac{1}{2}}{(3)}^{\frac{n-1}{2}}={(3)}^6 \\ \therefore \frac{1}{2}+\frac{n-1}{2}=6 \\ \Rightarrow \frac{1+n-1}{2}=6 \\ \Rightarrow n=12 \end{gathered}$$

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

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

Clearly this sequence is in G.P

With $a=\frac{1}{3}$ and $r=\frac{1}{9}÷\frac{1}{3}=\frac{1}{3}$

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

$$\begin{gathered} a_n={ar}^{n-1}\Rightarrow {ar}^{n-1}=\frac{1}{19683} \\ \Rightarrow \left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^{n-1}=\frac{1}{19683} \\ \Rightarrow {\left(\frac{1}{3}\right)}^n={\left(\frac{1}{3}\right)}^9 \\ \Rightarrow n=9 \end{gathered}$$

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