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 - Test - Page 871: 8



Work Step by Step

Here, $a_1=7, r=-2, n=10$ Now, we have $S_n=\dfrac{a_1(1-r^n)}{1-r}$ Now, $S_{10}=\dfrac{7(1-(-2)^{10})}{1-(-2)}$ or, $=-2387$
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