Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Mid-Chapter Check Point - Page 858: 10

Answer

$-29300$

Work Step by Step

Here $a_1=4 $ and $d=-6$; an arithmetic sequence $S_n=\dfrac{n}{2}(2a_1+(n-1)d)$ Thus, $S_{100}=\dfrac{100}{2}(2(4)+(100-1)(-6))=50(8+(-594))$ or, $=-29300$
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