Answer
$a_n=6\left(\dfrac{5}{3}\right)^{n-1}$
Work Step by Step
Let's note the geometric sequence by $\{a_n\}$ and the common ratio by $r$. We are given:
$$\begin{cases}
a_2=10\\
a_5=\dfrac{1250}{27}.
\end{cases}$$
The formula of the $n$th term is:
$$a_n=a_1r^{n-1}.\tag1$$
First we determine $r$ by dividing $a_5$ by $a_2$:
$$\begin{align*}
\dfrac{a_5}{a_2}&=\dfrac{a_1r^4}{a_1r}=r^3=\dfrac{\dfrac{1250}{27}}{10}=\dfrac{125}{27}\\
r^3&=\dfrac{125}{27}\\
r&=\sqrt[3]{\dfrac{125}{27}}\\
&=\dfrac{5}{3}.
\end{align*}$$
Then we determine $a_1$ using $r$ and $a_2$:
$$\begin{align*}
a_2&=a_1r\\
10&=a_1\cdot \dfrac{5}{3}\\
a_1&=6.
\end{align*}$$
Our sequence is:
$$\begin{cases}
a_1=6\\
r=\dfrac{5}{3}.
\end{cases}$$
We determine the $n$th using Eq. $(1)$:
$$a_n=6\left(\dfrac{5}{3}\right)^{n-1}.$$
So the $n$th term is $6\left(\dfrac{5}{3}\right)^{n-1}$.
