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 814: 16


$a_n=6(3)^{n-1}$ and $a_7=4374$

Work Step by Step

Here, we have $6,18,54, 162$ This shows a geometric series with a common ratio of $6r=18 $ Thus, $r=3$ The nth term is given by $a_n= a_1 r^{n-1}$ This gives: $a_n=(6) \times (3)^{n-1}$ Plug in $n=7$ $a_7=6 \times (3)^{7-1}=4374$ Hence, $a_n=6(3)^{n-1}$ and $a_7=4374$
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