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: 34

Answer

$a_n=6561\left(\frac{1}{3}\right)^{n-1}$

Work Step by Step

$a_6=27$ and $a_{12}=\frac{1}{27}$ $\frac{a_{12}}{a_6}=\frac{ar^{11}}{ar^5}=\frac{\frac{1}{27}}{27}=\frac{1}{729}=r^6$ $\implies r=\left(\frac{1}{729}\right)^{\frac{1}{6}}=\frac{1}{3}.$ $a_6=ar^5=a\left(\frac{1}{3}\right)^5=a\left(\frac{1}{243}\right)=27$ $\implies a=6561$ The $n^{th}$ term is therefore $a_n=6561\left(\frac{1}{3}\right)^{n-1}$
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