College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.3 - Geometric Sequences; Geometric Series - 9.3 Assess Your Understanding: 11

Answer

The sequence is geometric with a common ratio of $\frac{1}{2}$. The first four terms are: $a_1=-\frac{3}{2}$ $a_2=-\frac{3}{4}$ $a_3=-\frac{3}{8}$ $a_4=-\frac{3}{16}$

Work Step by Step

Substitute $1, 2, 3,$ and $4$ for $n$ into the given formula: $a_1=-3(\frac{1}{2})^1 = -3(\frac{1}{2})=-\frac{3}{2}$ $a_2=-3(\frac{1}{2})^2 = -3(\frac{1}{4})=-\frac{3}{4}$ $a_3=-3(\frac{1}{2})^3 = -3(\frac{1}{8})=-\frac{3}{8}$ $a_4=-3(\frac{1}{2})^4 = -3(\frac{1}{16})=-\frac{3}{16}$ Notice the the next term is equal to $\frac{1}{2}$ times the current term. This means that a common ratio of $\frac{1}{2}$ exists, and the sequence is geometric.
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