Answer
$6820$
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, $S_{10}=6820$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
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
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