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: 55

Answer

$\dfrac {3}{20}$

Work Step by Step

$r=\dfrac {\dfrac {-2}{27}}{\dfrac {1}{9}}=\dfrac {\dfrac {1}{9}}{\dfrac {-1}{6}}=\dfrac {-\dfrac {1}{6}}{\dfrac {1}{4}}=-\dfrac {2}{3}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {1}{4}}{1-\left( -\dfrac {2}{3}\right) }=\dfrac {3}{20}$
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