Question

Which term of the $GP2,2\sqrt{2},\mathrm{4..........}$ is $256$.

Answer 1

$a=2,r=\frac{2\sqrt{2}}{2}=\sqrt{2}$

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

$256=2(\sqrt{2}{)}^{n-1}$

$128=(\sqrt{2}{)}^{n-1}$

$2^7=2^{\frac{n-1}{2}}$

$7=\frac{n-1}{2}$

$14=n-1$

$\therefore n=15$

256 is the 15th term