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: 20

Answer

$a_n=-80\left(\frac{1}{4}\right)^{n-1}$

Work Step by Step

$-80,-20,-5,-\frac{5}{4},-\frac{5}{16},-\frac{5}{64},...$ $\frac{a_2}{a_1}=\frac{-20}{-80}=\frac{1}{4}=r$ $ra_2=\frac{1}{4}\times(-20)=-5=a_3$ $ra_3=\frac{1}{4}\times(-5)=-\frac{5}{4}=a_4$ and so on. The sequence is geometric with ratio $r=\frac{1}{4}$. The first term is $a=-80$ and so the $n^{th}$ term is given by $a_n=-80\left(\frac{1}{4}\right)^{n-1}$
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