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: 29


$a_n=x+3n-3$ $x_8=x+21$

Work Step by Step

$a_1=x$ $a_2=x+3$ Then $d=a_2-a_1=(x+3)-x=3$. $a_n=a_1+(n-1)d$ $a_n=x+(n-1)*3$ $\boxed{a_n=x+3n-3}$ $a_8=x+3*8-3$ $a_8=x+24-3$ $\boxed{x_8=x+21}$
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