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: 45


for $r=\frac{2}{3}$: $a_1=9$ $a_2=6$ $a_3=4$ $a_4=\frac{8}{3}$ $a_5=\frac{16}{9}$ for $r=\frac{2}{3}$: $a_1=9$ $a_2=-6$ $a_3=4$ $a_4=-\frac{8}{3}$ $a_5=\frac{16}{9}$

Work Step by Step

$a_1=9,~~a_3=4$ The nth term of a geometric sequence: $a_n=a_1r^{n-1}$ $a_3=a_1r^{3-1}$ $4=9r^2$ $\frac{4}{9}=r^2$ $r=\frac{2}{3}$ or $r=-\frac{2}{3}$ for $r=\frac{2}{3}$: $a_2=a_1+r^{2-1}=9(\frac{2}{3})=6$ $a_3=4$ $a_4=a_1+r^{4-1}=9(\frac{2}{3})^3=9(\frac{8}{27})=\frac{8}{3}$ $a_5=a_1+r^{5-1}=9(\frac{2}{3})^4=9(\frac{16}{81})=\frac{16}{9}$ for $r=\frac{2}{3}$: $a_2=a_1+r^{2-1}=9(-\frac{2}{3})=-6$ $a_3=4$ $a_4=a_1+r^{4-1}=9(-\frac{2}{3})^3=9(-\frac{8}{27})=-\frac{8}{3}$ $a_5=a_1+r^{5-1}=9(-\frac{2}{3})^4=9(\frac{16}{81})=\frac{16}{9}$
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