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: 5

Answer

$155$

Work Step by Step

$a_{1}=5;r=\dfrac {40}{20}=\dfrac {20}{10}=\dfrac {10}{5}=2\Rightarrow S_{n}=\dfrac {a_{1}\left( 1-r^{n}\right) }{1-r}=\dfrac {5\times \left( 1-2^{5}\right) }{1-2}=155$
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