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 - Page 736: 3

Answer

Geometric with $r=\frac{1}{3}$

Work Step by Step

$270, 90, 30, 10, \frac{10}{3} ,\frac{10}{9} , . . .$ $\frac{a_2}{a_1}=\frac{90}{270}=\frac{1}{3}=r$ $ra_2=30=a_3$ $ra_3=10=a_4$ $ra_4=\frac{10}{3}=a_5$ and so on. The sequence is geometric with common ratio $r=\frac{1}{3}$.
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