Algebra and Trigonometry 10th Edition

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

Chapter 11 - Review Exercises - Page 840: 33

Answer

80

Work Step by Step

Need to find $S = \sum_{n=1}^{10}(2J-3) = (\sum_{n=1}^{10}2J)-(\sum_{n=1}^{10}3) = 2(\sum_{n=1}^{10}J) - 30$ (Since we know from arithmetic sequence sum that $\sum_{i=1}^{n}J = \frac{n(n+1)}{2})$. Hence, $S = 2(\sum_{n=1}^{10}J) - 30 = 2(\frac{10(10+1)}{2})) - 30 = 110 - 30 = 80$
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