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: 19


$a_n=2 (\dfrac{3}{4})^{n-1}$ and $a_7=\dfrac{729}{2048}$

Work Step by Step

The nth term is given by $a_n= a_1 r^{n-1}$ Here, $a_1=2; r=\dfrac{a_2}{a_1}=\dfrac{3/2}{2}=\dfrac{3}{4}$ $a_n=(2) \times (\dfrac{3}{4})^{n-1}$ This gives: $a_n=2 (\dfrac{3}{4})^{n-1}$ Plug in $n=7$. $a_7=2 (\dfrac{3}{4})^{7-1}=\dfrac{729}{2048}$ Hence, $a_n=2 (\dfrac{3}{4})^{n-1}$ and $a_7=\dfrac{729}{2048}$
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