Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Review Exercises - Page 1125: 46



Work Step by Step

The sum of a infinite geometric sequence is given as: $ S=\dfrac{a_1}{1-r}$ Here, $ a_1=9$ and Common Ratio: $ r=\dfrac{1}{3}$ $ S=\dfrac{a_1}{1-r}$ Now, $ S=\dfrac{9}{1-\dfrac{1}{3}}=\dfrac{27}{3-1}=\frac{27}{2}=13.5$
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