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 11.2 Arithmetic Sequences - 11.2 Assess Your Understanding - Page 834: 25


$a_{90} =-266$

Work Step by Step

The $n^{th}$ term of an arithmetic sequence is given by the formula: $a_n = a_1 + (n-1)d (1)$ where $a_1 = \ First \ Term; \\ d = \ Common \ Difference$ We have: $a_1=1 \\ d=a_2-a_1=-2-1=-3$ We will substitute the above data into formula (1) to obtain: $a_n=1+(-3) (n-1) \implies a_n=1-3(n-1)$ In order to compute the $90th$ term, we need to plug in $90$ for $n$ into the above form to obtain: $a_{90} = 1-3(90-1)=1-(89)(3)$ So, $a_{90} =-266$
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