Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - Review Exercises - Page 841: 44


$a_1=6$ $a_2=-2$ $a_3=\frac{2}{3}$ $a_4=-\frac{2}{9}$ $a_5=\frac{2}{27}$

Work Step by Step

The nth term of a geometric sequence: $a_n=a_1r^{n-1}$ $a_1=6,~~r=-\frac{1}{3}$ $a_2=6(-\frac{1}{3})^{2-1}=6(-\frac{1}{3})=-2$ $a_3=6(-\frac{1}{3})^{3-1}=6(\frac{1}{9})=\frac{6}{9}=\frac{2}{3}$ $a_4=6(-\frac{1}{3})^{4-1}=6(-\frac{1}{27})=-\frac{6}{27}=-\frac{2}{9}$ $a_5=6(-\frac{1}{3})^{5-1}=6(\frac{1}{81})=\frac{6}{81}=\frac{2}{27}$
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