## College Algebra (11th Edition)

Published by Pearson

# Chapter 7 - Section 7.3 - Geometric Sequences and Series - 7.3 Exercises - Page 653: 8

#### Answer

$a_5=-32 \text{ and } a_n=-\dfrac{1}{8}\cdot4^{n-1}$

#### 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_3=-2 \\ r=4 ,\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 $a_n=a_1r^{n-1},$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\ a_3=a_1r^{3-1} \\\\ a_3=a_1r^{2} .\end{array} Subtituting the given values, $a_3=-2$ and $r=4,$ the equation above results to \begin{array}{l}\require{cancel} -2=a_1(4^{2}) \\\\ -2=a_1(16) \\\\ -\dfrac{2}{16}=a_1 \\\\ a_1=-\dfrac{1}{8} .\end{array} With $a_1=-\dfrac{1}{8}$ and $r=4,$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\ a_n=-\dfrac{1}{8}\cdot4^{n-1} .\end{array} With $n=5,$ the equation above bcomes \begin{array}{l}\require{cancel} a_5=-\dfrac{1}{8}\cdot4^{5-1} \\\\ a_5=-\dfrac{1}{8}\cdot4^{4} \\\\ a_5=-\dfrac{1}{8}\cdot256 \\\\ a_5=-1(32) \\\\ a_5=-32 .\end{array} Hence, $a_5=-32 \text{ and } a_n=-\dfrac{1}{8}\cdot4^{n-1} .$

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