## Calculus with Applications (10th Edition)

$${a_1} = 27,{a_2} = 9,{a_3} = 3,{a_4} = 1$$
\eqalign{ & {a_2} = 9,\,\,\,\,{a_3} = 3,\,\,\,\,n = 4 \cr & {\text{find }}r,{\text{ using }}r = \frac{{{a_{n + 1}}}}{{{a_n}}} \cr & r = \frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_3}}}{{{a_2}}} = \frac{3}{9} \cr & r = \frac{1}{3} \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} = 9,\,\,\,\,r = \frac{1}{3}{\text{ into }}{a_n} = a{r^{n - 1}}{\text{,}}\,\,\,\,\,n = 2{\text{ to obtain}} \cr & 9 = {a_1}{\left( {\frac{1}{3}} \right)^{2 - 1}} \cr & 9 = {a_1}\left( {\frac{1}{3}} \right) \cr & {a_1} = 27 \cr & {\text{the general term is }}{a_n} = 27{\left( {\frac{1}{3}} \right)^{n - 1}} \cr & \cr & {\text{list the }}n = 4{\text{ terms}}{\text{. then}} \cr & {a_1} = 27 \cr & {a_2} = 9 \cr & {a_3} = 27{\left( {\frac{1}{3}} \right)^{3 - 1}} = 3 \cr & {a_4} = 27{\left( {\frac{1}{3}} \right)^{4 - 1}} = 1 \cr & \cr & {\text{then}} \cr & {a_1} = 27,{a_2} = 9,{a_3} = 3,{a_4} = 1 \cr}