## College Algebra (11th Edition)

Published by Pearson

# Chapter 7 - Section 7.2 - Arithmetic Sequences and Series - 7.2 Exercises: 20

#### Answer

$a_8=y-17; \\a_n=(y+4)+(n-1)(-3)$

#### 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_2=y+1, d=-3 ,\end{array} use the formula for finding the $n$th term of an arithmetic sequence. $\bf{\text{Solution Details:}}$ Using $a_n=a_1+(n-1)d$ with $a_{2}=y+1,$ $d=-3,$ and $n=2$ then \begin{array}{l}\require{cancel} a_n=a_1+(n-1)d \\\\ a_2=a_1+(2-1)d \\\\ a_2=a_1+(1)d \\\\ a_2=a_1+d \\\\ y+1=a_1+(-3) \\\\ y+1=a_1-3 \\\\ y+1+3=a_1 \\\\ y+4=a_1 \\\\ a_1=y+4 .\end{array} Using $a_n=a_1+(n-1)d$ with $a_{1}=y+4,$ and $d=-3,$ then \begin{array}{l}\require{cancel} a_n=a_1+(n-1)d \\\\ a_n=y+4+(n-1)(-3) .\end{array} With $n=8,$ then \begin{array}{l}\require{cancel} a_8=y+4+(8-1)(-3) \\\\ a_8=y+4+(7)(-3) \\\\ a_8=y+4-21 \\\\ a_8=y-17 .\end{array} Hence, $a_8=y-17 \text{ and } a_n=y+4+(n-1)(-3) .$

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