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 1061: 82


200th position

Work Step by Step

The sum of an arithmetic sequence is given by: $S_n=\dfrac{n}{2}[a_1+a_n]$ and the nth term for an arithmetic sequence is given by $a_n=a_1+(n-1) d$ We are given that $a_n= 314, 628$ Now, $314, 628=21,700+(n-1) \times 1472$ or, $n-1= \dfrac{292, 928}{1472}$ or, $n=200$ Hence, the given number is at the 200th position in the given sequence.
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