Answer
The sequence is geometric with a common ratio of $\frac{5}{2}$
The first four terms are:
$b_1=\frac{5}{2}$
$b_2=\frac{25}{4}$
$b_3=\frac{125}{8}$
$b_4=\frac{625}{16}$
Work Step by Step
Substitute $1, 2, 3,$ and $4$ for $n$ into the given formula:
$b_1=(\frac{5}{2})^1 = \frac{5}{2}$
$b_2=(\frac{5}{2})^2 = \frac{5^2}{2^2}=\frac{25}{4}$
$b_3=(\frac{5}{2})^3 = \frac{5^3}{2^3}=\frac{125}{8}$
$b_4=(\frac{5}{2})^4 = \frac{5^4}{2^4}=\frac{625}{16}$
Notice the next term is equal to $\frac{5}{2}$ times the current term.
This means that a common ratio of $\frac{5}{2}$ exists, and the sequence is geometric.