Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.1 - Sequences - Exercise Set: 33

Answer

$a_{n}=4n-1$

Work Step by Step

Given Sequence: $3,7,11,15$ We can write it as $4-1,8-1,12-1,16-1$ $ a_{1} = 3 = 4-1 = 4(1)-1$ $ a_{2} = 7 = 8-1 = 4(2)-1$ $ a_{3} = 11 = 12-1 = 4(3)-1$ $ a_{4} = 15 = 16-1 = 4(4)-1$ General term : $a_{n}=4(n)-1$
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