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 - Mid-Chapter Check Point - Page 858: 5

Answer

$1536$

Work Step by Step

Here $a_1=3 $ and $r=2$; a geometric sequence $a_n=a_1r^{n-1}=3(2)^{n-1}$; Thus, $a_{10}=3(2)^{10-1}=3(2)^9$ or, $a_{10}=1536$
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