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: 40


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

Work Step by Step

The $n^{th}$ term of the 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 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: $a_3 = a_1 (r)(r); \\a_3 = a_1 \ (1/3)^2 \ \\ \dfrac{1}{3}=\dfrac{a_1}{9} \implies a_1=3$ Therefore, the $n^{th}$ term of the geometric sequence is: $a_n=3 (\dfrac{1}{3})^{n-1}$
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