Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.3 Geometric Sequences and Series - 11.3 Exercises - Page 1035: 53

Answer

$27$

Work Step by Step

$r=\dfrac {\dfrac {2}{3}}{2}=\dfrac {2}{6}=\dfrac {6}{18}=\dfrac {1}{3}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {18}{1-\dfrac {1}{3}}=27$
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