Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Mid-Chapter Check Point - Page 1077: 6



Work Step by Step

we have $a_1=3 $ and $r=2$ For a geometric sequence $a_n=a_1r^{n-1}$ Thus, $a_n=3(2)^{n-1}$ Now, $a_{10}=3(2)^{10-1}=3(2)^9=1536$
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