## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.1 Geometric Sequences - 12.1 Exercises - Page 612: 16

#### Answer

$$r = 4{\text{ and }}{a_n} = {4^n}$$

#### Work Step by Step

\eqalign{ & {a_1} = 4,{a_2} = 16,{a_3} = 64,{a_4} = 256,... \cr & {\text{to verify that the sequence is geometric}}{\text{, divide each term except the first by }} \cr & {\text{the preceding term}} \cr & \frac{{256}}{{64}} = \frac{{64}}{{16}} = \frac{{16}}{4} = 4 \cr & {\text{the ratio is constant}}{\text{, ao the sequence is geometric with }}r = 4 \cr & {\text{The general term of a geometric sequence is }}{a_n} = {a_1}{r^{n - 1}} \cr & {\text{then}} \cr & {a_n} = 4{\left( 4 \right)^{n - 1}} \cr & {\text{simplifying}} \cr & {a_n} = {4^n} \cr}

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