Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 14 - Sequences, Series, and the Binomial Theorem - Mid-Chapter Review - Mixed Review - Page 915: 12

Answer

$S_{\infty}=\frac{250}{3}\approx 83.33$

Work Step by Step

In order to determine whether the infinite geometric series has a limit we have to find $r$. If $|r| < 1$ the it has a limit. Our geometric series has the terms: $100, -20, 4,\cdots $. We calculate the ratio: $r=\frac{a_2}{a_1}=\frac{-20}{100}=-\frac{1}{5}$ Because $|r|=\left|-\frac{1}{5}\right|=\frac{1}{5}<1$, it follows that the series has a limit. Let's find it. The limit of an infinite geometric series is $S_{\infty}=\dfrac{a_1}{1-r}$. In our case $$S_{\infty}=\dfrac{100}{1-\left(-\frac{1}{5}\right)}=\dfrac{100}{\frac{6}{5}}=\frac{500}{6}=\frac{250}{3}\approx 83.33$$
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