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: 30

Answer

$r=\dfrac {1}{2};a_{1}=-2$

Work Step by Step

$\dfrac {a_{m}}{a_{n}}=r^{m-n}\Rightarrow \dfrac {a_{g}}{a_{4}}=r^{5}=\dfrac {-\dfrac {1}{128}}{-\dfrac {1}{4}}=\dfrac {1}{32}\Rightarrow r=\dfrac {1}{2};a_{4}=a_{1}\times r^{3}\Rightarrow -\dfrac {1}{4}=a_{1}\times \left( \dfrac {1}{2}\right) ^{3}\Rightarrow a_{1}=-2$
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