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: 64

Answer

$\dfrac {5}{4}=1.25$

Work Step by Step

$\sum ^{\infty }_{i=1}\left( \dfrac {5}{9}\right) ^{i}=\dfrac {a_{1}}{1-r};a_{1}=\left( \dfrac {5}{9}\right) ^{1}=\dfrac {5}{9};r=\dfrac {5}{9}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {5}{9}}{1-\dfrac {5}{9}}=\dfrac {5}{4}=1.25$
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