## Calculus with Applications (10th Edition)

$${a_1} = \frac{1}{2},{a_2} = 2,{a_3} = 8,{a_4} = 32$$
\eqalign{ & {a_1} = \frac{1}{2},\,\,\,\,r = 4,\,\,\,\,n = 4 \cr & {\text{The general term of a geometric sequence is }}{a_n} = a{r^{n - 1}}.{\text{ with first term }}a{\text{ and }} \cr & {\text{common ratio }}r. \cr & {\text{substituting }}{a_1} = \frac{1}{2},\,\,\,\,r = 4{\text{ into }}{a_n} = a{r^{n - 1}}{\text{ to obtain}} \cr & {a_n} = \left( {\frac{1}{2}} \right){\left( 4 \right)^{n - 1}} \cr & {\text{list the }}n = 4{\text{ terms}}{\text{. then}} \cr & {a_1} = \frac{1}{2} \cr & {a_2} = \left( {\frac{1}{2}} \right){\left( 4 \right)^{2 - 1}} = \left( {\frac{1}{2}} \right){\left( 4 \right)^{2 - 1}} = 2 \cr & {a_3} = \left( {\frac{1}{2}} \right){\left( 4 \right)^{3 - 1}} = \left( {\frac{1}{2}} \right){\left( 4 \right)^2} = 8 \cr & {a_4} = \left( {\frac{1}{2}} \right){\left( 4 \right)^{4 - 1}} = \left( {\frac{1}{2}} \right){\left( 4 \right)^3} = 32 \cr & \cr & {\text{then}} \cr & {a_1} = \frac{1}{2},{a_2} = 2,{a_3} = 8,{a_4} = 32 \cr}