Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.3 Geometric Sequences; Geometric Series - 11.3 Assess Your Understanding - Page 844: 11


$a_1=-\dfrac{3}{2}$ $a_2=-\dfrac{3}{4}$ $a_3=-\dfrac{3}{8}$ $a_4=-\dfrac{3}{16}$ The sequence is geometric with a common ratio of $\dfrac{1}{2}$.

Work Step by Step

We need to substitute $1, 2, 3,$ and $4$ for $n$ into the given equation to find the first four terms. $a_1=-3(\dfrac{1}{2})^1 = -3(\dfrac{1}{2})=-\dfrac{3}{2}$ $a_2=-3(\dfrac{1}{2})^2 = -3(\dfrac{1}{4})=-\dfrac{3}{4}$ $a_3=-3(\dfrac{1}{2})^3 = -3(\dfrac{1}{8})=-\dfrac{3}{8}$ $a_4=-3(\dfrac{1}{2})^4 = -3(\dfrac{1}{16})=-\dfrac{3}{16}$ We see that the the next term is equal to $\dfrac{1}{2}$ times the current term and this implies that the sequence is geometric with a common ratio of $\dfrac{1}{2}$.
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