College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 7 - Section 7.3 - Geometric Sequences and Series - 7.3 Exercises - Page 653: 6

Answer

$a_5=5,000 \text{ and } a_n=8(-5)^{n-1}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find $a_5$ and $a_n$ for the given geometric sequence described as \begin{array}{l}\require{cancel} a_1=8 \\ r=-5 ,\end{array} use the formula for finding the $n^{th}$ term of a geometric sequence. $\bf{\text{Solution Details:}}$ Using the formula for finding the $n^{th}$ term of a geometric sequence, which is given by $a_n=a_1r^{n-1}$ with $a_1=8$ and $r=-5,$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\ a_n=8(-5)^{n-1} .\end{array} With $n=5,$ then \begin{array}{l}\require{cancel} a_5=8(-5)^{5-1} \\\\ a_5=8(-5)^{4} \\\\ a_5=8(625) \\\\ a_5=5,000 .\end{array} Hence, $ a_5=5,000 \text{ and } a_n=8(-5)^{n-1} .$
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