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: 35

Answer

$\dfrac {99}{8}$

Work Step by Step

$r=\dfrac {\dfrac {-9}{4}}{\dfrac {9}{2}}=\dfrac {\dfrac {9}{2}}{-9}=\dfrac {-9}{18}=-\dfrac {1}{2};a_{1}=18\Rightarrow S_{n}=\dfrac {a_{1}\left( r^{n}-1\right) }{r-1}=\dfrac {18\times \left( \left( \dfrac {-1}{2}\right) ^{5}-1\right) }{-\dfrac {1}{2}-1}=\dfrac {18\times \left( \dfrac {-33}{32}\right) }{-\dfrac {3}{2}}=\dfrac {99}{8}$
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