Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - Mixed Review of Problem Solving - Lessons 12.1-12.3 - Page 818: 7b


$a_n=66 (\dfrac{1}{2})^{n-1}$

Work Step by Step

From the previous part it can be seen that $a_1=66; r=\dfrac{1}{2}$. Thus, the series is geometric. So, we have $a_n=a_1r^{n-1}$ But $a_1=66; r=\dfrac{1}{2}$ Then $a_n=66 (\dfrac{1}{2})^{n-1}$
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