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: 20


$a_n=3 (\dfrac{-2}{5})^{n-1}$ and $a_7=\dfrac{192}{15625}$

Work Step by Step

The nth term is given by $a_n= a_1 r^{n-1}$ Here, $a_1=3; r=\dfrac{a_2}{a_1}=\dfrac{-6/5}{2}=\dfrac{-2}{5}$ $a_n=(3) \times (\dfrac{-2}{5})^{n-1}$ This gives: $a_n=3 (\dfrac{-2}{5})^{n-1}$ Plug in $n=7$ $a_7=3 (\dfrac{-2}{5})^{7-1}=\dfrac{192}{15625}$ Hence, $a_n=3 (\dfrac{-2}{5})^{n-1}$ and $a_7=\dfrac{192}{15625}$
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