Algebra and Trigonometry 10th Edition

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

Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 795: 41

Answer

$a_9=\frac{1}{768}$

Work Step by Step

$a_1=\frac{1}{3}$ $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}$ $r=\frac{-\frac{1}{6}}{\frac{1}{3}}=\frac{\frac{1}{12}}{-\frac{1}{6}}=-\frac{1}{2}$ $a_n=a_1r^{n-1}$ $a_9=\frac{1}{3}(-\frac{1}{2})^{9-1}=\frac{1}{3}(-\frac{1}{2})^8=\frac{1}{3}(\frac{1}{256})=\frac{1}{768}$
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