Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Mixed Review - Page 825: 56


$a_n=\dfrac{10240}{81} (0.75)^{n-1}= 126.42 (0.75)^{n-1} $

Work Step by Step

Here, we have $a_n= a_1r^{(n-1)}$ for the Geometric series. We are told that $a_6=30$ and $r=0.75$ Here, we have $a_6=(a_1) \times (0.75)^{(6-1)} \implies 30=(a_1) \times (0.75)^5$ or, $a_1=\dfrac{30}{0.2373}=126.42$ Now, $a_n=a_1 \times r^{n-1}$ Hence, $a_n=\dfrac{10240}{81} (0.75)^{n-1}= 126.42 (0.75)^{n-1} $
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