Question

Find the last term of the following sequence $4,4^2,4^3$ ............. to $2n$ terms

Answer 1

It is given that the first term of G.P is $a_1=4$ and the second term is $a_2=4^2$, therefore,

$r=\frac{a_2}{a_1}=\frac{4^2}{4}=4$

We know that the formula for the $n$th term of an G.P is $T_n=a{r}^{n-1}$, where $a$ is the first term, $r$ is the common ratio.

It is given that the first term is $a=4$, number of terms are $n=2n$ and the common ratio is $r=4$, thus,

$T_{2n}=4\times {\left(4\right)}^{2n-1}=4\times 4^{2n}\times 4^{-1}=4\times (4^2{)}^n\times \frac{1}{4}={16}^n$

Hence, the last term is${16}^n$, which is equivalent to $4^{2n}$.