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: 44


$a_n=6 (-4)^{n-1}$

Work Step by Step

The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) Here, we have $a_2=a_1 r$ and $a_5=a_1 r^{5-1} \implies a_5=a_1 r^4$ $\dfrac{a_5}{a_2}=\dfrac{a_1r^4}{a_1r}$ This gives: $r^3=-\sqrt {64} \implies r=-4$ Now, $a_2=a_1 r \implies a_1= \dfrac{-24}{-4}=6$ Hence, $a_n=6 (-4)^{n-1}$
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