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: 25

Answer

$a_{5}=a_{1}\times r^{4}=10\times \left( -\dfrac {1}{2}\right) ^{4}=\dfrac {-5}{8};$ $a_{n}=a_{1}\times \left( r^{n-1}\right) =10\times \left( -\dfrac {1}{2}\right) ^{n-1}$

Work Step by Step

$r=\dfrac {\dfrac {-5}{4}}{\dfrac {5}{2}}=\dfrac {\dfrac {5}{2}}{-5}=\dfrac {-5}{10}=-\dfrac {1}{2};a_{1}=10\Rightarrow a_{5}=a_{1}\times r^{4}=10\times \left( -\dfrac {1}{2}\right) ^{4}=\dfrac {-5}{8};a_{n}=a_{1}\times \left( r^{n-1}\right) =10\times \left( -\dfrac {1}{2}\right) ^{n-1}$
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