College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Review Exercises - Page 788: 12

Answer

See below.

Work Step by Step

$a_n=7+4(n-1)$, thus $a_1=7+4(1-1)=7+4\cdot0=7+0=7$ $a_2=7+4(2-1)=7+4\cdot1=7+4=11$ $a_3=7+4(3-1)=7+4\cdot2=7+8=15$ $a_4=7+4(4-1)=7+4\cdot3=7+12=19$ $a_5=7+4(5-1)=7+4\cdot4=7+16=23$ $a_6=7+4(6-1)=7+4\cdot5=7+20=27$
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