Question

Find the $8th$ term of the geometric progression $2,\frac{2}{3},\frac{2}{9},\frac{2}{27},...$

Answer 1

For a GP, the $n^{th}$ term is given by:

$a_n=a{r}^{n-1}$

$⟹a_8=a{r}^7$

Given that: $a=2$ and $r=\frac{1}{3}$

We get, $a_8=2\times {\left(\frac{1}{3}\right)}^7$

$a_8=\frac{2}{3^7}$ $\text{Ans.}$