Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 716: 21

Answer

converges and its sum is $\frac{400}{9}$

Work Step by Step

In this explicit form we see that $r$ =|$0.73|\lt1 $ which means that this geometric series will converge. Knowing that $a$ refers to the first term which is $12$ and using the explicit $r$, we get $S= \frac{a}{1-r}$ = $\frac{12}{1-0.73} = \frac{12}{0.27} = \frac{400}{9}$ Hence, the series converges and its sum is $\frac{400}{9}$ .
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