Question

Find the${10}^{\mathrm{th}}$ term of the GP : $12,4,\frac{4}{3},\frac{4}{9}$.

Answer 1

Find the ${10}^{\mathrm{th}}$ term of GP:

Here, the sequence is $12,4,\frac{4}{3},\frac{4}{9}$

First term $a=12$,

Common difference $r=\frac{4}{12}=\frac{1}{3}$

The general term of GP is $a_n=a{r}^{n-1}$

$\begin{array}{rcl}& \Rightarrow & a_{10}=12\times {\left(\frac{1}{3}\right)}^{10-1}\\ & =& 12\times {\left(\frac{1}{3}\right)}^9\\ & =& \frac{4}{6561}\end{array}$

Hence, then the term of given GP is $\frac{4}{6561}$