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: 38

Answer

$a_n=390625\left(-\frac{2}{5}\right)^{n-1}$

Work Step by Step

$a_7=1600$ and $a_{12}=-16.384$ $\frac{a_{12}}{a_7}=\frac{ar^{11}}{ar^{6}}=r^5=\frac{-16.384}{1600}=-\frac{32}{3125}$ $\implies r=\left(-\frac{32}{3125}\right)^{\frac{1}{5}}=-\frac{2}{5}$ $a_7=ar^6=a\left(-\frac{2}{5}\right)^{6}=a\frac{64}{15625}=1600$ $\implies a=390625$ The $n^{th}$ term is therefore given by $a_n=390625\left(-\frac{2}{5}\right)^{n-1}$
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