Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.3 Geometric Sequences; Geometric Series - 12.3 Assess Your Understanding - Page 824: 35



Work Step by Step

We know that $a_n=a_1r^{n-1}.$ where $r$=common ration (we can get this by dividing any two consecutive terms) and $a_1$= the first term Here $a_1=-3$ and $r=\frac{a_2}{a_1}=\dfrac{1}{-3}=-\frac{1}{3}$ Hence, here we have $a_n=-3(-\frac{1}{3})^{n-1}.$
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