Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Sections 11.1-11.3 - Integrated Review - Sequences and Series - Page 653: 14

Answer

$a_{6}=\dfrac{243}{16}$

Work Step by Step

Using $a_n=a_1r^{n-1}$, then \begin{array}{l} a_{6}=64\left( \dfrac{3}{4} \right)^{6-1}\\\\ a_{6}=64\left( \dfrac{243}{1024} \right)\\\\ a_{6}=\dfrac{243}{16} .\end{array}
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