Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - Summary Exercises on Sequences and Series - Exercises: 5

Answer

Geometric sequence $r=\dfrac {4}{3}$

Work Step by Step

$\dfrac {\dfrac {64}{27}}{\dfrac {16}{9}}=\dfrac {\dfrac {16}{9}}{\dfrac {4}{3}}=\dfrac {\dfrac {4}{3}}{1}=\dfrac {1}{\dfrac {3}{4}}=\dfrac {4}{3}\Rightarrow r=\dfrac {4}{3}$ Then this is geometric sequence
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