Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.2 Arithmetic Sequences and Series - 11.2 Exercises - Page 1024: 24

Answer

$a_n=1-4n$ $a_8=-31$

Work Step by Step

$-3, -7, -11, ...$ We have $a_1=-3$ and $d=-7-(-3)=-4$. Using the formula $a_n=a_1+(n-1)d$, we get: $a_n=-3+(n-1)*(-4)$ $a_n=-3-4n+4$ $\boxed{a_n=1-4n}$ Then $a_8=1-4*8=1-32=\boxed{-31}$.
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