Question

Find the $7^{th}$ term of the GP, whose first term is $6$ and common ration is $\frac{2}{3}$.

Answer 1

Given that, $t_1=6$ and r = 2/3
We have $t_n=a.r^{n-1}$
$\therefore t_7=\left(6\right){\left(\frac{2}{3}\right)}^6=\frac{\left(6\right)\left(64\right)}{729}=\frac{128}{243}$