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: 52

Answer

$\dfrac {3}{2}$

Work Step by Step

$r=\dfrac {\dfrac {1}{27}}{\dfrac {1}{9}}=\dfrac {\dfrac {1}{9}}{\dfrac {1}{3}}=\dfrac {\dfrac {1}{3}}{1}=\dfrac {1}{3};a_{1}=1\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {1}{1-\dfrac {1}{3}}=\dfrac {3}{2}$
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