Answer
$2350$
Work Step by Step
we have $a_1=-2 $ and $d=2$
For an arithmetic sequence, we have $S_n=\dfrac{n}{2}(2a_1+(n-1)d)$
Now, $S_{50}=\dfrac{50}{2}(2(-2)+(50-1)2)$
or, $=25(-4+98)$
Thus, $S_{50}=2350$
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: 9
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