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 737: 33

Answer

$a_n=16384\left(\frac{1}{4}\right)^{n-1}$

Work Step by Step

$a_4=256$ and $a_{10}=\frac{1}{16}$ $\frac{a_{10}}{a_4}=\frac{ar^{10}}{ar^4}=\frac{\frac{1}{16}}{256}=\frac{1}{4096}=r^6$ $\implies r=(\frac{1}{4096})^{\frac{1}{6}}=\frac{1}{4}.$ $a_4=ar^3=a\left(\frac{1}{4}\right)^3=a\left(\frac{1}{64}\right)=256$ $\implies a=16384$ The $n^{th}$ term is therefore $a_n=16384\left(\frac{1}{4}\right)^{n-1}$
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