Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set: 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} $ $a_{4} = 3 \times (-\frac{8}{27})$ $a_{4} = -\frac{8}{9}$
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