Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.2 - Arithmetic Sequences - Exercise Set - Page 1060: 55



Work Step by Step

Here, we have $a_{14}=1+(14-1)(-3-1)=-51$ and $b_{14}=3+(14-1)(8-3)=68$ The sum of an arithmetic sequence is given by: $S_n=\dfrac{n}{2}[a_1+a_n]$ Now, $\sum_{i=1}^{14}b_i-\sum_{i=1}^{14} a_i=\dfrac{14}{2}(3+68)-\dfrac{14}{2}(1-51)$ Thus, $\sum_{i=1}^{14}b_i-\sum_{i=1}^{14} a_i=847$
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