## College Algebra (10th Edition)

The sequence is geometric with a common ratio of $\frac{5}{2}$ The first four terms are: $b_1=\frac{5}{2}$ $b_2=\frac{25}{4}$ $b_3=\frac{125}{8}$ $b_4=\frac{625}{16}$
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.