Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.1 - Sequences and Series - 11.1 Exercises - Page 779: 97c


$A_{80}\ne 2\times A_{40}$

Work Step by Step

$A_n=10,000(1+\frac{0.035}{4})^n$ In order to find the balance after 20 years we need to compute the 80th term of the sequence: $A_{80}=10,000(1+\frac{0.035}{4})^{80}=10,000(1.00875)^{80}=20076.31$ From item (b): $A_{40}=14,169.09$ It is clear that $A_{80}\ne 2\times A_{40}=28,338.18$
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