# Chapter 12 - Sequences and Series - 12.1 Geometric Sequences - 12.1 Exercises - Page 612: 8

$${a_n} = 8{\left( 4 \right)^{n - 1}}{\text{ and }}{a_5} = 2048$$

#### Work Step by Step

\eqalign{ & {a_1} = 8,\,\,\,\,r = 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} = a = 8,\,\,\,\,r = 4{\text{ into }}{a_n} = a{r^{n - 1}}{\text{ to obtain}} \cr & {a_n} = {a_1}{r^{n - 1}} = 8{\left( 4 \right)^{n - 1}} \cr & \cr & {\text{find }}{a_5},{\text{ substitute }}n = 5{\text{ into the general term formula}} \cr & {a_5} = 8{\left( 4 \right)^{5 - 1}} \cr & {a_5} = 8{\left( 4 \right)^4} \cr & {a_5} = 2048 \cr & {\text{then}} \cr & {a_n} = 8{\left( 4 \right)^{n - 1}}{\text{ and }}{a_5} = 2048 \cr}

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