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: 41



Work Step by Step

The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) When $n=4$, then we have Equation (1) gives: $a_4=a_1(r)^{4-1}=a_1r^3$ Plug in $a_4=-16$ and $a_1=-\dfrac{1}{4}$ $-16=-\dfrac{1}{4}r^3$ This gives: $r=4$ Hence, $a_n=-\dfrac{1}{4}(4)^{n-1}$
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