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: 39

Answer

See below

Work Step by Step

The common ratio $r$, can be found as $$r^2 = 12/3 = 4$$ Thus $r= 2, -2$ So the nth term, $a_n = (2)^{n-1}*3$ or $(-2)^{n-1}*3$
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