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: 10



Work Step by Step

we have: $a_1=-20 $ and $r=-2$ For a geometric sequence, we have $S_n=\dfrac{a_1(1-r^n)}{1-r}$ Now, $S_{10}=\dfrac{-20(1-(-2)^{10})}{1-2}$ or, $=\dfrac{20460}{3}$ Thus, $S_{10}=6820$
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