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: 67


The answer is $2280$.

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 $10$ terms. $S_{10}=\frac{-5(1-(-2)^{10})}{1-(-2)}$. Simplify. $S_{10}=1705$. And the second series is ${b_n=10,-5,-20,-35,...}$. Common difference $d=-15$. The tenth terms is. $T_{10}=10+(10-1)(-15)$ Simplify. $T_{10}=10+9(-15)$ $T_{10}=10-135$ $T_{10}=-125$ The sum of the first $10$ terms is $S_{10}=\frac{10}{2}\cdot (10-125)$ $S_{10}=5\cdot (-115)$ $S_{10}=-575$. The difference is $=1705-(-575)$ $=2280$.
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