Answer
$225$
Work Step by Step
Here, $a_1=-6, d=3$
Sum of first $15$th term of an arithmetic sequence can be calculated as:
$S_n=\dfrac{n}{2}[2a_1+(n-1)d]$
Now,
$S_{15}=\dfrac{15}{2}[2(-6)+(15-1)3]=\dfrac{15}{2}[-12+(14)3]=225$
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13417-894-7
ISBN 13:
978-0-13417-894-3
Chapter 11 - Review Exercises - Page 869: 20
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