College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Review Exercises - Page 789: 46

Answer

$13.5$

Work Step by Step

We can see that the ratio of subsequent terms is $\frac{1}{3}$, thus $r=\frac{1}{3}$. The sum of an infinte geometric series can be dividing the first term($a_1$) by the difference of $1$ and the common ratio ($r$). So $S=\frac{a_1}{1-r}$. Hence here $S=\frac{9}{1-\frac{1}{3}}=13.5$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.