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 - Page 1034: 23

Answer

$a_{5}=a_{1}\times r^{4}=\dfrac {4}{5}\times \left( \dfrac {5}{2}\right) ^{4}=\dfrac {125}{4};$ $a_{n}=a_{1}\times r^{n-1}=\dfrac {4}{5}\times \left( \dfrac {2}{5}\right) ^{n-1}$

Work Step by Step

$r=\dfrac {\dfrac {25}{2}}{5}=\dfrac {5}{2}=\dfrac {2}{\dfrac {4}{5}}=\dfrac {5}{2};a_{1}=\dfrac {4}{5}\Rightarrow a_{5}=a_{1}\times r^{4}=\dfrac {4}{5}\times \left( \dfrac {5}{2}\right) ^{4}=\dfrac {125}{4};a_{n}=a_{1}\times r^{n-1}=\dfrac {4}{5}\times \left( \dfrac {2}{5}\right) ^{n-1}$
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