Intermediate Algebra (6th Edition)

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

Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set - Page 647: 51

Answer

$a_{n}=6(3)^{n-1}$

Work Step by Step

Given, $a_{1}=6$ $r = 3$ Substituting $a_{1}$ and $r$ in $a_{n}= a_{1}(r)^{n-1}$ we get, $a_{n}=6(3)^{n-1}$
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