Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Review Exercises - Page 870: 33



Work Step by Step

Since, we have $a_n=a_1r^{n-1}$ Here, $n=6,a_1=16,r=\dfrac{1}{2}$ Thus, $a_6=(16)(\dfrac{1}{2})^{6-1}=16(\dfrac{1}{32})=\dfrac{1}{2}$
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