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: 43

Answer

$a_{15}=\dfrac{17}{2}$

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_{15}=\dfrac{3}{2}+(15-1)\left( \dfrac{1}{2} \right) \\\\ a_{15}=\dfrac{3}{2}+7 \\\\ a_{15}=\dfrac{17}{2} .\end{array}
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