Question

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

• A. 3
• B. 9
• C. 6
• D. None of these

Answer 1

Answer: (B)

$a=\frac{1}{3},r=\frac{1/9}{1/3}=\frac{1}{9}\times \frac{3}{1}=\frac{1}{3}$
Let $T_n=\frac{1}{19683}$
$\Rightarrow a{r}^{n-1}=\frac{1}{19683}$
$\Rightarrow \frac{1}{3}{\left(\frac{1}{3}\right)}^{n-1}=\frac{1}{19683}$
$\Rightarrow {\left(\frac{1}{3}\right)}^n={\left(\frac{1}{3}\right)}^9$
$\Rightarrow n=9$