Intermediate Algebra (6th Edition)

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

Chapter 11 - Review - Page 668: 45

Answer

$-10$

Work Step by Step

Using the given general term, $ a_n=-8+(n-1)3 ,$ then, \begin{array}{l} a_1=-8+(1-1)3\\ a_1=-8+0\\ a_1=-8 ,\\\\ a_2=-8+(2-1)3\\ a_2=-8+3\\ a_2=-5 ,\\\\ a_3=-8+(3-1)3\\ a_3=-8+6\\ a_3=-2 ,\\\\ a_4=-8+(4-1)3\\ a_4=-8+9\\ a_4=1 ,\\\\ a_5=-8+(5-1)3\\ a_5=-8+12\\ a_5=4 .\end{array} Hence, $ S_5=-8+(-5)+(-2)+1+4 = -10 .$
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