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: 36

Answer

$\dfrac {244}{27}$

Work Step by Step

$r=\dfrac {\dfrac {-4}{9}}{\dfrac {4}{3}}=\dfrac {\dfrac {4}{3}}{-4}=\dfrac {-a}{12}=-\dfrac {1}{3};a_{1}=12\Rightarrow S_{n}=\dfrac {a_{1}\left( r^{n}-1\right) }{r-1}=\dfrac {12\times \left( \left( -\dfrac {1}{3}\right) ^{5}-1\right) }{\left( -\dfrac {1}{3}-1\right) }=\dfrac {12\times \left( \dfrac {-244}{243}\right) }{\dfrac {-4}{3}}=\dfrac {244}{27}$
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