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: 10

Answer

$a_5=36 \text{ and } a_n=\dfrac{9}{4}(2^{n-1})$

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=18 \\ r=2 ,\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, $a_4=18$ and $r=2,$ the equation above results to \begin{array}{l}\require{cancel} 18=a_1(2^{3}) \\\\ 18=a_1(8) \\\\ \dfrac{18}{8}=a_1 \\\\ a_1=\dfrac{9}{4} .\end{array} With $a_1=\dfrac{9}{4}$ and $r=2,$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\ a_n=\dfrac{9}{4}(2^{n-1}) .\end{array} With $n=5,$ the equation above bcomes \begin{array}{l}\require{cancel} a_5=\dfrac{9}{4}(2^{5-1}) \\\\ a_5=\dfrac{9}{4}(2^{4}) \\\\ a_5=\dfrac{9}{4}(16) \\\\ a_5=9(4) \\\\ a_5=36 .\end{array} Hence, $ a_5=36 \text{ and } a_n=\dfrac{9}{4}(2^{n-1}) .$
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