Question

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 $\frac{1}{19683}?$

Answer 1

(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 $128$.

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

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 $729$.

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

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$\frac{1}{19683}$

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

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