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: 40

Answer

$a_{5} = 648$

Work Step by Step

$a_{1} = 8$ $r = -3$ $a_{n} $ of geometric sequence is $a_{n} = a_{1} . r^{n-1} $ To find fifth term, substitute $a_{1} ,$ $r$ values and $n=5$ $a_{5} = a_{1} . r^{5-1} $ $a_{5} = a_{1} . r^{4} $ $a_{5} = 8 \times(-3)^{4} $ $a_{5} = 8 \times 81 $ $a_{5} = 648$
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