Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Review - Page 668: 54

Answer

$S_{100}=-30,000$

Work Step by Step

Using $S_n=\dfrac{n}{2}(2a_1+(n-1)d)$, with $a_1=-3$ and $d=-6$, then, \begin{array}{l} S_{100}=\dfrac{100}{2}(2(-3)+(100-1)(-6))\\\\ S_{100}=50(-6+(-594))\\\\\ S_{100}=50(-600)\\\\\ S_{100}=-30,000 .\end{array}
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