Answer
$S_{40}=-7020$
Work Step by Step
$d=-9-0=-18-(-9)=-9$.
$a_n=a_1+(n-1)d$
$a_{40}=0+(40-1)(-9)$
$a_{40}=39(-9)$
$a_{40}=-351$
$S_n=\frac{n}{2}(a_1+a_n)$
$S_{40}=\frac{40}{2}(0-351)=20(-351)=-7020$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 11 - 11.2 - Arithmetic Sequences and Partial Sums - 11.2 Exercises - Page 786: 53
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