College Algebra (6th Edition)

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

Chapter 8 - Mid-Chapter Check Point - Page 743: 4

Answer

First term (put n = 1) $a_{1}$ = 3 Second term (put n = 2) $a_{2}$ = 1 Third term (put n = 3) $a_{3}$ = 3 Fourth term (put n = 4) $a_{4}$ = 1 Fifth term (put n = 5) $a_{5}$ = 3

Work Step by Step

Given $a_{1}$ = 3, $a_{n}$ = $-a_{(n - 1)}$ + 4 First term (put n = 1) $a_{1}$ = 3 Second term (put n = 2) $a_{2}$ = $-a_{(2 - 1)}$ + 4 = $-a_{1}$ + 4 = -3 + 4 = 1 Third term (put n = 3) $a_{3}$ = $-a_{(3 - 1)}$ + 4 = $-a_{2}$ + 4 = -1 + 4 = 3 Fourth term (put n = 4) $a_{4}$ = $-a_{(4 - 1)}$ + 4 = $-a_{3}$ + 4 = -3 + 4 = 1 Fifth term (put n = 5) $a_{5}$ = $-a_{(5 - 1)}$ + 4 = $-a_{4}$ + 4 = -1 + 4 = 3
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