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: 35

Answer

$a_n=(\dfrac{-1}{3})^{n-2}$

Work Step by Step

The $n^{th}$ term of the geometric sequence is given by the formula: $ a_n=a_1r^{n-1}$ where $r$=common ratio and $a_1$= the first term The common ratio of a geometric sequence is equal to the quotient (ratio) of any term and the term before it: $ \ r = \dfrac{a_n}{a_{n-1}}$ or, $r=\dfrac{a_2}{a_1}$ Here, we have: $a_1= -3$ and $a_2=1$, so $r=\dfrac{1}{-3}=-3$ Thus, the general formula for the given sequence is: $a_n=-3(\dfrac{-1}{3})^{n-1}=(\dfrac{-1}{3})^{n-2}$
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