Question

If a = 1 and r $=\frac{2}{3}$, find (a) $T_n$ (b) $T_4$

Answer 1

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 $a=1$ and the common ratio is $r=\frac{2}{3}$, thus,

(a) $T_n=1\times {\left(\frac{2}{3}\right)}^{n-1}={\left(\frac{2}{3}\right)}^{n-1}$

(b) $T_4=1\times {\left(\frac{2}{3}\right)}^{4-1}={\left(\frac{2}{3}\right)}^3=\frac{2^3}{3^3}=\frac{8}{27}$

Hence, $T_n={\left(\frac{2}{3}\right)}^{n-1}$ and $T_4=\frac{8}{27}$.