## College Algebra (11th Edition)

Published by Pearson

# Chapter 7 - Section 7.3 - Geometric Sequences and Series - 7.3 Exercises - Page 653: 9

#### Answer

$a_5=-729 \text{ and } a_n=-9(-3)^{n-1}$

#### 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_4=243 \\ r=-3 ,\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_4=a_1r^{4-1} \\\\ a_4=a_1r^{3} .\end{array} Subtituting the given values, $a_4=243$ and $r=-3,$ the equation above results to \begin{array}{l}\require{cancel} 243=a_1(-3)^{3} \\\\ 243=a_1(-27) \\\\ \dfrac{243}{-27}=a_1 \\\\ a_1=-9 .\end{array} With $a_1=-9$ and $r=-3,$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\ a_n=-9(-3)^{n-1} .\end{array} With $n=5,$ the equation above bcomes \begin{array}{l}\require{cancel} a_5=-9(-3)^{5-1} \\\\ a_5=-9(-3)^{4} \\\\ a_5=-9(81) \\\\ a_5=-729 .\end{array} Hence, $a_5=-729 \text{ and } a_n=-9(-3)^{n-1} .$

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