Question

Which term of the following sequences :

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

Answer 1

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 $\frac{1}{19683}$