Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Skill Practice - Page 815: 45


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

Work Step by Step

The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) Here, we have $162=a_1 r^3$ and $4374=(162/r^3) (r^6/1) \implies r=3$ $a_1=\dfrac{162}{(3)^3} \implies a_1=6$ Equation (1) becomes: $a_n=6 (3)^{n-1}$ Hence, $a_n=6 (3)^{n-1}$
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