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 - Section 11.3 - Geometric Sequences and Series - Exercise Set - Page 855: 70

Answer

The answer is $1140$.

Work Step by Step

For the first series ${a_n}=-5,10,-20,40,...$. Common ratio $r=-2$. This is the geometric series. Sum of the first $9$ terms. $S_{9}=\frac{-5(1-(-2)^{9})}{1-(-2)}$. Simplify. $S_{9}=-855$. The second series is $c_n=-2,1,-\frac{1}{2},\frac{1}{4}...$ Common ratio $r=-\frac{1}{2}$ This is the geometric series. Sum of the infinite terms is $S_{\infty}=\frac{-2}{1-\left ( -\frac{1}{2} \right)}$ $S_{\infty}=\frac{-2}{1+\frac{1}{2} }$ $S_{\infty}=\frac{-2}{\frac{2+1}{2} }$ $S_{\infty}=-\frac{4}{3}$ The product is $=(-855)\cdot (-\frac{4}{3})$. $=1140$.
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