## Precalculus (10th Edition)

In order for a sequence to be geometric, the quotient of all consecutive terms must be constant. Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{(\frac{5}{2})^{n+1}}{(\frac{5}{2})^n}=\frac{5}{2}$, thus it is a geometric sequence. $a_1=\frac{5}{2}$ $a_2=\frac{25}{4}$ $a_3=\frac{125}{8}$ $a_4=\frac{625}{16}$
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant. Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{(\frac{5}{2})^{n+1}}{(\frac{5}{2})^n}=\frac{5}{2}$, thus it is a geometric sequence. $a_1=(\frac{5}{2})^1=\frac{5}{2}$ $a_2=(\frac{5}{2})^2=\frac{25}{4}$ $a_3=(\frac{5}{2})^3=\frac{125}{8}$ $a_4=(\frac{5}{2})^4=\frac{625}{16}$