Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 9 - Conic Sections, Sequences, and Series - 9.4 Geometric Sequences - 9.4 Exercises: 32

Answer

$a_n=9(2)^{n-1}$

Work Step by Step

$a_6=288$ and $a_{13}=36864$ $\frac{a_{13}}{a_6}=\frac{ar^12}{ar^5}=\frac{36864}{288}=128=r^7$ $\implies r=(128)^{\frac{1}{7}}=2.$ $a_6=ar^5=a(2^5)=32a=288$ $\implies a=9$ The $n^{th}$ term is therefore $a_n=9(2)^{n-1}$
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