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: 16

Answer

$a_n=200+(n-1)(-20)$ ; $a_{20}=-180$

Work Step by Step

Here, $a_1=200, d=-20$ n-th term an arithmetic series can be calculated as: $a_n=a_1+(n-1)d$ Thus, $a_n=200+(n-1)(-20)$ Also, $a_{20}=200+(20-1)(-20)=-180$
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