Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.2 - Arithmetic Sequences - Exercise Set - Page 842: 95



Work Step by Step

Let us consider the sequence $a_n=a_13^{n-1}$ Now, $n=7$ and $a_1=11$ Thus, $a_7=a_13^{7-1}$ or, $a_7=a_13^{6}$ or, $a_7=(11)(729)=8019$ Hence, $a_7=8019$
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.