College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Test - Page 791: 10

Answer

$\frac{1}{4^9}$

Work Step by Step

We can see that the ratio of subsequent terms is $\frac{1}{4}$, thus $r=\frac{1}{4}$. The nth term of a geometric series can be obtained by the following formula: $a_n=a_1\cdot r^{n-1}$, where $a_1$ is the first term and $r$ is the common ratio. Hence here: $a_{12}=16\cdot(\frac{1}{4})^{12-1}=\frac{1}{4^9}$
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