Question

The $4^{th}$ term of a geometric sequence is $\frac{2}{3}$ and the seventh term is $\frac{16}{81}$. Find the geometric sequence

Answer 1

Given that $t_4=\frac{2}{3}$ and $t_7=\frac{16}{81}$.
Using the formula $t_n=a{r}^{n-1},n=1,2,3,....$ for the general term we have,
$t_4=a{r}^3=\frac{2}{3}$ and $t_7=a{r}^6=\frac{16}{81}$.
Note that in order to find the geometric sequence, we need to find $a$ and $r$.
By dividing $t_7$ by $t_4$ we obtain,
$\frac{t_7}{t_4}=\frac{a{r}^6}{a{r}^3}=\frac{\frac{16}{81}}{\frac{2}{3}}=\frac{8}{27}$.
Thus, $r^3=\frac{8}{27}={\left(\frac{2}{3}\right)}^3$ which implies $r=\frac{2}{3}$.
Now, $t_4=\frac{2}{3}\Rightarrow a{r}^3=\left(\frac{2}{3}\right)$.
$\Rightarrow a\left(\frac{8}{27}\right)=\frac{2}{3}$

$\Rightarrow a=\frac{9}{4}$
Hence, the required geometric sequence is $a,ar,a{r}^2,a{r}^3,...,a{r}^{n-1},a{r}^n,....$
That is, $\frac{9}{4},\frac{9}{4}\left(\frac{2}{3}\right),\frac{9}{4}{\left(\frac{2}{3}\right)}^2,.....$