## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set - Page 646: 41

#### Answer

$a_{4}= - \frac{8}{9}$

#### Work Step by Step

Given, $a_{1} = 3$ $r = -\frac{2}{3}$ $a_{n}$ of geometric sequence is $a_{n} = a_{1} . r^{n-1}$ To find fourth term, substitute $a_{1} ,$ $r$ values and $n=4$ $a_{4} = a_{1} . r^{4-1}$ $a_{4} = a_{1} . r^{3}$ $a_{4} = 3 \times (-\frac{2}{3})^{3}$ $a_{4} = 3 \times (-\frac{8}{27})$ $a_{4} = -\frac{8}{9}$

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.