Answer
$a_n=y-3n+7$
$a_8=y-17$
Work Step by Step
$a_2=y+1$
$d=-3$
Find $a_1$:
$a_2=a_1+d$
$y+1=a_1+(-3)$
$a_1=y+4$
$a_n=a_1+(n-1)d$
$a_n=(y+4)+(n-1)*(-3)$
$a_n=y+4-3n+3$
$\boxed{a_n=y-3n+7}$
$a_8=y-3*8+7$
$a_8=y-24+7$
$\boxed{a_8=y-17}$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
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: 30
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