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: 57

Answer

$20301$

Work Step by Step

$a_{1}+a_{2}+a_{3}\ldots +a_{n}=\dfrac {a_{1}+a_{n}}{2}\times n$ $d=a_{n+1}-a_{n}=9-5=5-1=4$ $a_{n}=a_{1}+\left( n-1\right) d\Rightarrow 401=1+\left( n-1\right) \times 4\Rightarrow \left( n-1\right) =\dfrac {401-1}{4}=100\Rightarrow n=101$ $1+5+9\ldots +401=\dfrac {1+401}{2}\times 101=20301$
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