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 1034: 51



Work Step by Step

$r=\dfrac {a_{n+1}}{a_{n}}=\dfrac {\dfrac {1}{4}}{\dfrac {1}{2}}=\dfrac {\dfrac {1}{2}}{1}=\dfrac {1}{2}=\dfrac {1}{2};a_{1}=2\Rightarrow S_{n}=\dfrac {a_{1}}{1-r}=\dfrac {2}{1-\dfrac {1}{2}}=\dfrac {2}{\dfrac {1}{2}}=4$
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