College Algebra (11th Edition)

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

Chapter 7 - Section 7.2 - Arithmetic Sequences and Series - 7.2 Exercises: 18

Answer

$a_8=29 \text{ and } a_n=-3n+53$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find $a_8$ and $a_n$ in the given sequence \begin{array}{l}\require{cancel} a_{15}=8, a_{16}=5 ,\end{array} use the formula for finding the $n$th term of an arithmetic sequence. $\bf{\text{Solution Details:}}$ Since the common difference, $d,$ is the difference between a term and the term preceeding it, then \begin{array}{l}\require{cancel} d=a_{16}-a_{15} \\\\ d=5-8 \\\\ d=-3 .\end{array} Using $a_n=a_1+(n-1)d$ with $a_{15}=8$ $d=-3,$ and $n=15,$ then \begin{array}{l}\require{cancel} a_n=a_1+(n-1)d \\\\ a_{15}=a_1+(15-1)d \\\\ 8=a_1+(14)(-3) \\\\ 8=a_1-42 \\\\ 8+42=a_1 \\\\ a_1=50 .\end{array} Using $a_n=a_1+(n-1)d$ with $a_1=50$ and $d=-3,$ then \begin{array}{l}\require{cancel} a_n=50+(n-1)(-3) \\\\ a_n=50-3n+3 \\\\ a_n=-3n+53 .\end{array} With $n=8,$ then \begin{array}{l}\require{cancel} a_8=-3(8)+53 \\\\ a_8=-24+53 \\\\ a_8=29 .\end{array} Hence, $ a_8=29 \text{ and } a_n=-3n+53 .$
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