College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises: 11

Answer

$a_{n}=\frac{5}{2}(-\frac{1}{2})^{n-1}$ $a_{4}=-\frac{5}{16}$

Work Step by Step

We are given: $a= \frac{5}{2}, r=- \frac{1}{2}$ We know that a geometric sequence has the form: $a_{n}=ar^{n-1}$ Thus: $a_{n}=\frac{5}{2}(-\frac{1}{2})^{n-1}$ And so: $a_{4}= \frac{5}{2}*(-\frac{1}{2})^{3}=-\frac{5}{16}$
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