Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.2 Arithmetic Sequences and Series - 11.2 Exercises - Page 1026: 72

Answer

-765

Work Step by Step

$\sum ^{19}_{k=5}\left( -3-4k\right) =\dfrac {\left( a_{5}+a_{19}\right) \times n}{2}\Rightarrow a_{5}=-3-4\times 5=-23;a_{19}=-3-4\times 19=-79\Rightarrow S=\dfrac {15\times \left( -23+\left( -79\right) \right) }{2}=15\times \left( -51\right) =-765$
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