Answer
$a_{10}=101$
Work Step by Step
Using $a_n=a_1+(n-1)d$, then
\begin{array}{l}
a_{10}=20+(10-1)9\\
a_{10}=20+(9)9\\
a_{10}=20+81\\
a_{10}=101
.\end{array}
Intermediate Algebra (6th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321785045
ISBN 13:
978-0-32178-504-6
Chapter 11 - Sections 11.1-11.3 - Integrated Review - Sequences and Series - Page 653: 13
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