## Calculus with Applications (10th Edition)

$$r = 2{\text{ and }}{a_n} = \frac{3}{4}{\left( {2} \right)^{n - 1}}$$
\eqalign{ & {a_1} = \frac{3}{4},{a_2} = \frac{3}{2},{a_3} = 3,{a_4} = 6,{a_5} = 12,... \cr & {\text{to verify that the sequence is geometric}}{\text{, divide each term except the first by }} \cr & {\text{the preceding term}} \cr & \frac{{12}}{6} = \frac{6}{3} = \frac{3}{{3/2}} = \frac{{3/2}}{{3/4}} = 2 \cr & {\text{}}{\text{ the ratio is constant}}{\text{, so the sequence is geometric with }}r = 2 \cr & {\text{the general term of a geometric sequence is }}{a_n} = {a_1}{r^{n - 1}} \cr & {\text{then}} \cr & {a_n} = \frac{3}{4}{\left( {2} \right)^{n - 1}} \cr}