Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.3 Geometric Sequences and Series - 11.3 Exercises: 18

Answer

$a_{5}=a_{1}\times r^{5-1}=a_{1}\times r^{4}=\left( -\dfrac {1}{8}\right) \times 4^{4}=-32;$ $a_{n}=a_{1}\times r^{n-1}=\left( -\dfrac {1}{8}\right) \times 4^{n-1}$

Work Step by Step

$a_{1}\times r^{2}=a_{3}\Rightarrow a_{1}=\dfrac {a_{3}}{r^{2}}=\dfrac {-2}{4^{2}}=-\dfrac {1}{8};a_{5}=a_{1}\times r^{5-1}=a_{1}\times r^{4}=\left( -\dfrac {1}{8}\right) \times 4^{4}=-32;a_{n}=a_{1}\times r^{n-1}=\left( -\dfrac {1}{8}\right) \times 4^{n-1}$
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