Question

Which term of the sequence $\frac{1}{3},\frac{1}{9},\frac{1}{27},....$ is $\frac{1}{19683}$ ?

Answer 1

$\frac{1}{3},\frac{1}{9},\frac{1}{27},...$

We know that $a_n=a{r}^{n-1}$

First term$a=\frac{1}{3}$

Common ratio$r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}$

$n^{th}$ term$=\frac{1}{19683}$

Substituting these values we have

$n^{th}$ term$=a{r}^{n-1}$

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

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

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

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

Comparing the powers,we get

$n=9$

Hence $9^{th}$ term of G.P is $\frac{1}{19683}$