Precalculus (6th Edition) Blitzer

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

Chapter 10 - Mid-Chapter Check Point - Page 1077: 7


$-\frac{1}{2}n+2$, $-13$

Work Step by Step

Step 1. With the given numbers, we can identify the series as arithemetic. Thus we have $a_1=\frac{3}{2}, d=-\frac{1}{2}$ and $a_n=a_1+(n-1)d=\frac{3}{2}-\frac{1}{2}(n-1)=-\frac{1}{2}n+2$. Step 2. We have $a_{30}=-\frac{1}{2}(30)+2=-13$
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