## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.1 Geometric Sequences - 12.1 Exercises - Page 612: 22

#### Answer

$$r = - \frac{1}{3}{\text{ and }}{a_n} = \frac{7}{4}{\left( { - \frac{1}{3}} \right)^{n - 1}}$$

#### Work Step by Step

\eqalign{ & {a_1} = \frac{7}{4},{a_2} = - \frac{7}{{12}},{a_3} = \frac{7}{{36}},{a_4} = - \frac{7}{{108}},... \cr & {\text{to verify that the sequence is geometric}}{\text{, divide each term except the first by }} \cr & {\text{the preceding term}} \cr & \frac{{{a_4}}}{{{a_3}}} = \frac{{ - 7/108}}{{7/36}} = - \frac{1}{3} \cr & \frac{{{a_3}}}{{{a_2}}} = \frac{{7/36}}{{ - 7/12}} = - \frac{1}{3} \cr & \frac{{{a_2}}}{{{a_1}}} = \frac{{ - 7/12}}{{7/4}} = - \frac{1}{3} \cr & {\text{the ratio is constant}}{\text{, so the sequence is geometric with }}r = - \frac{1}{3} \cr & {\text{the general term of a geometric sequence is }}{a_n} = {a_1}{r^{n - 1}} \cr & {\text{then}} \cr & {a_n} = \frac{7}{4}{\left( { - \frac{1}{3}} \right)^{n - 1}} \cr}

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