Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set - Page 647: 44

Answer

$a_{11}=-\dfrac{4}{3}$

Work Step by Step

Using $a_n=a_1+(n-1)d$ or the $n$th term of an arithmetic sequence, then, \begin{array}{l} a_{11}=2+(11-1)\left( -\dfrac{1}{3} \right) \\\\ a_{11}=2-\dfrac{10}{3} \\\\ a_{11}=\dfrac{6-10}{3} \\\\ a_{11}=\dfrac{-4}{3} \\\\ a_{11}=-\dfrac{4}{3} .\end{array}
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