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: 37

Answer

$a_n=256\left(-\frac{1}{2}\right)^{n-1}$

Work Step by Step

$a_5=16$ and $a_{12}=-\frac{1}{8}$ $\frac{a_{12}}{a_5}=\frac{ar^{11}}{ar^{4}}=r^7=\frac{-\frac{1}{8}}{16}=-\frac{1}{128}$ $\implies r=\left(-\frac{1}{128}\right)^{\frac{1}{7}}=-\frac{1}{2}$ $a_5=ar^4=a\left(-\frac{1}{2}\right)^{4}=a\frac{1}{16}=16$ $\implies a=256$ The $n^{th}$ term is therefore given by $a_n=256\left(-\frac{1}{2}\right)^{n-1}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.