Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.2 - Arithmetic Sequences - Exercise Set - Page 1060: 59



Work Step by Step

We know that $a_n=a_1+(n-1) d$ Here, we have $a_{2}=a_1+d=4; a_{6}=a_1+5d=16$ Thus,we have $a_1+5d-a_1-d=16-12$ or, $4d=12 \implies d=3$ and $a_1=4-d=4-3=1$ Now, $a_n=1+(n-1) (3)=3n-2$
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