Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.2 - Arithmetic Sequences - 12.2 Exercises - Page 857: 60



Work Step by Step

See p. 855. For the arithmetic sequence $a_{n}=a+(n-1)d$ the nth partial sum $S_{n}=\displaystyle \sum_{k=1}^{n}[a+(k-1)d]$ is given by either of the following equivalent formulas: 1. $S_{n}=\displaystyle \frac{n}{2}[2a+(n-1)d]\qquad $2. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$ ------------------- We see from the given sequence that $a=89$ $d=85-89=-4$ The last term is $a_{n}=13$ (we find n): $a_{n}=a+(n-1)d$ $13=89-4(n-1)\qquad/-89$ $-76=-4(n-1)\qquad/\div(-4)$ $19=n-1$ $20=n$ And, we find $S_{20}$, (using formula 2) $S_{20}=20(\displaystyle \frac{89+13}{2})=10(102)=1020$
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