College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Chapter 8 Review - Exercises - Page 640: 21

Answer

Geometric $a_5=\frac{4}{27}$

Work Step by Step

We are given the sequence $\{a_n\}$: $$\dfrac{3}{4}, \dfrac{1}{2},\dfrac{1}{3},\dfrac{2}{9}\dots $$ Check if the sequence is arithmetic: $$\begin{align*} a_2-a_1&=\dfrac{1}{2}-\dfrac{3}{4}=-\dfrac{1}{4}\\ a_3-a_2&=\dfrac{1}{3}-\dfrac{1}{2}=-\dfrac{1}{6}\\ a_4-a_3&=\dfrac{2}{9}-\dfrac{1}{3}=-\dfrac{1}{9}. \end{align*}$$ Since the differences between consecutive terms are not constant, the sequence is not arithmetic. Check if the sequence is geometric: $$\begin{align*} \dfrac{a_2}{a_1}&=\dfrac{\frac{1}{2}}{\frac{3}{4}}=\dfrac{2}{3}\\ \dfrac{a_3}{a_2}&=\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{2}{3}\\ \dfrac{a_4}{a_3}&=\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{3}. \end{align*}$$ Since the ratios between consecutive terms are constant, the sequence is geometric. Its first term is $a_1=\frac{3}{4}$ and its common ratio $r=\frac{2}{3}$. We calculate the fifth term: $$a_5=a_4r=\dfrac{2}{9}\cdot\dfrac{2}{3}=\dfrac{4}{27}.$$
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.