Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 795: 43

Answer

$a_3=9$

Work Step by Step

$a_1=16$ $a_n=a_1r^{n-1}$ $a_4=16r^{4-1}$ $\frac{27}{4}=16r^3$ $r^3=\frac{27}{64}$ $r=\sqrt[3] {\frac{27}{64}}=\frac{3}{4}$ $a_3=16(\frac{3}{4})^{3-1}=16·\frac{9}{16}=9$
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