Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 9 - Conic Sections, Sequences, and Series - 9.3 Arithmetic Sequences - 9.3 Exercises: 17

Answer

$n=14$

Work Step by Step

$a_n=2n^2+5$ when $a_n=397$ $397=2n^2+5$ $\implies 392=2n^2$ $\implies 196=n^2$ $\implies \sqrt{196}=n$ $\implies n=14$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.