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: 7


$a_5=-108 \text{ and } a_n=-\dfrac{4}{3}\cdot3^{n-1}$

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_2=-4 \\ 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_2=a_1r^{2-1} \\\\ a_2=a_1r^{1} \\\\ a_2=a_1r .\end{array} Subtituting the given values, $a_2=-4$ and $r=3,$ the equation above results to \begin{array}{l}\require{cancel} -4=a_1(3) \\\\ -\dfrac{4}{3}=a_1 \\\\ a_1=-\dfrac{4}{3} .\end{array} With $a_1=-\dfrac{4}{3}$ and $r=3,$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\ a_n=-\dfrac{4}{3}\cdot3^{n-1} .\end{array} With $n=5,$ the equation above bcomes \begin{array}{l}\require{cancel} a_5=-\dfrac{4}{3}\cdot3^{5-1} \\\\ a_5=-\dfrac{4}{3}\cdot3^{4} \\\\ a_5=-\dfrac{4}{3}\cdot81 \\\\ a_5=-4(27) \\\\ a_5=-108 .\end{array} Hence, $ a_5=-108 \text{ and } a_n=-\dfrac{4}{3}\cdot3^{n-1} .$
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