Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Mid-Chapter Check Point - Page 1077: 11



Work Step by Step

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