## Calculus with Applications (10th Edition)

$${a_n} = 24{\left( {\frac{1}{2}} \right)^{n - 1}}{\text{ and }}{a_5} = \frac{3}{2}$$
\eqalign{ & {a_2} = 12,\,\,\,\,r = \frac{1}{2} \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_2} = 12\,\,\,\,r = \frac{1}{2}{\text{,}}\,\,\,n = 2{\text{ into }}{a_n} = a{r^{n - 1}}{\text{ to obtain}} \cr & 12 = {a_1}{\left( {\frac{1}{2}} \right)^{2 - 1}} \cr & 12 = {a_1}\left( {\frac{1}{2}} \right) \cr & {a_1} = 24 \cr & {\text{find the general term }}{a_n} = {a_1}{r^{n - 1}} \cr & {a_n} = 24{\left( {\frac{1}{2}} \right)^{n - 1}} \cr & \cr & {\text{find }}{a_5},{\text{ substitute }}n = 5{\text{ into the general term formula}} \cr & {a_5} = 24{\left( {\frac{1}{2}} \right)^{5 - 1}} \cr & {a_5} = 24{\left( {\frac{1}{2}} \right)^4} \cr & {a_5} = \frac{3}{2} \cr & {\text{then}} \cr & {a_n} = 24{\left( {\frac{1}{2}} \right)^{n - 1}}{\text{ and }}{a_5} = \frac{3}{2} \cr}