Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 9 - Conic Sections, Sequences, and Series - 9.4 Geometric Sequences - 9.4 Exercises: 12

Answer

$a_n=8(3)^{n-1}$

Work Step by Step

$ 8, 24, 72, 216, 648, 1944, . . .$ $\frac{a_2}{a_1}=\frac{24}{8}=3=r$ $ra_2=3\times24=72=a_3$ $ra_3=3\times72=216=a_4$ and so on. The sequence is geometric with common ratio $r=3$. The $n^{th}$ term of the sequence is given by $a_n=8r^{n-1}=8(3)^{n-1}$
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