College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 7 - Section 7.3 - Geometric Sequences and Series - 7.3 Exercises - Page 654: 43


The sum converges and equals to $\frac{3}{20}=0.15$.

Work Step by Step

In a geometric series the ratio between consecutive terms is constant. In order to find this ratio, we simply divide two consecutive terms: $r=\frac{a_2}{a_1}=\frac{-\frac{1}{6}}{\frac{1}{4}}=-\frac{4}{6}=-\frac{2}{3}$ The sum of an infinite geometric series exists only if $\mid r\mid<1$, here it is true. Therefore the sum converges. The sum of an infinite geometric series is: $S_\infty=\frac{a_1}{1-r}$ where $a_1$=first term and $r$=common ratio The given series has $a_1=\frac{1}{4}$ and $r=-\frac{2}{3}$. Thus, $S_\infty=\frac{\frac{1}{4}}{1-(-\frac{2}{3})}=\frac{\frac{1}{4}}{\frac{5}{3}}=\frac{3}{20}=0.15$
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