## Calculus with Applications (10th Edition)

$${a_1} = \frac{3}{2},{a_2} = 3,{a_3} = 6,{a_4} = 12,{a_5} = 24$$
\eqalign{ & {a_3} = 6,\,\,\,\,{a_4} = 12,\,\,\,\,n = 5 \cr & {\text{find }}r,{\text{ using }}r = \frac{{{a_{n + 1}}}}{{{a_n}}} \cr & r = \frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_4}}}{{{a_3}}} = \frac{{12}}{6} \cr & r = 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_3} = 6,\,\,\,\,r = 2{\text{ into }}{a_n} = a{r^{n - 1}}{\text{,}}\,\,\,\,\,n = 3{\text{ to obtain}} \cr & 6 = {a_1}{\left( 2 \right)^{3 - 1}} \cr & 6 = {a_1}\left( 4 \right) \cr & {a_1} = \frac{6}{4} = \frac{3}{2} \cr & {\text{the general term is }}{a_n} = \frac{3}{2}{\left( 2 \right)^{n - 1}} \cr & \cr & {\text{list the }}n = 5{\text{ terms}}{\text{. then}} \cr & {a_1} = \frac{3}{2} \cr & {a_2} = \left( {\frac{3}{2}} \right){\left( 2 \right)^{2 - 1}} = 3 \cr & {a_3} = 6 \cr & {a_4} = 12 \cr & {a_5} = \left( {\frac{3}{2}} \right){\left( 2 \right)^{5 - 1}} = 24 \cr & \cr & {\text{then}} \cr & {a_1} = \frac{3}{2},{a_2} = 3,{a_3} = 6,{a_4} = 12,{a_5} = 24 \cr}