Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Review Exercises - Page 869: 20



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, $S_{15}=\dfrac{15}{2}[2(-6)+(15-1)3]=\dfrac{15}{2}[-12+(14)3]=225$
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