Answer
$S=5$
Work Step by Step
$\displaystyle \sum_{n=0}^{∞}4(0.2)^n=4(0.2)^0+4(0.2)^1+4(0.2)^2+...$
$a_1=4(0.2)^0=4$
$r=\frac{a_2}{a_1}=\frac{4(0.2)^1}{4(0.2)^0}=0.2$
$S=\frac{a_1}{1-r}=\frac{4}{1-0.2}=\frac{4}{0.8}=5$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 796: 74
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