Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Sections 11.1-11.3 - Integrated Review - Sequences and Series - Page 653: 11

Answer

$\left\{ 45,15,5,\dfrac{5}{3},\dfrac{5}{9} \right\} $

Work Step by Step

With $a_1=45$ and $r=\dfrac{1}{3}$, then }\begin{array}{l} a_{1}=45 ,\\\\ a_2=a_1r\\ a_2=45\left( \dfrac{1}{3} \right)\\ a_2=15 ,\\\\ a_3=a_2r\\ a_3=15\left( \dfrac{1}{3} \right)\\ a_3=5 ,\\\\ a_4=a_3r\\ a_4=5\left( \dfrac{1}{3} \right)\\ a_4=\dfrac{5}{3} ,\\\\ a_5=a_4r\\ a_5=\dfrac{5}{3}\left( \dfrac{1}{3} \right)\\ a_5=\dfrac{5}{9} .\end{array} Hence, the first 5 terms are $ \left\{ 45,15,5,\dfrac{5}{3},\dfrac{5}{9} \right\} $
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