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 - Page 1038: 5


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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.