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: 16

Answer

$a_n=81\left(\frac{1}{3}\right)^{n-1}$

Work Step by Step

$81, 27, 9, 3, 1, \frac{1}{3} , . . . $ $\frac{a_2}{a_1}=\frac{27}{81}=\frac{1}{3}=r$ $ra_2=\frac{1}{3}\times27=9=a_3$ $ra_3=\frac{1}{3}\times9=3=a_4$ and so on. The sequence is geometric with ratio $r=\frac{1}{3}$. The $n^{th}$ term is given by $a_n=81\left(\frac{1}{3}\right)^{n-1}$
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