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: 50

Answer

125

Work Step by Step

$a_{4}=a_{3}+d\Rightarrow 8=5+d\Rightarrow d=3\Rightarrow a_{3}=a_{1}+2d\Rightarrow 5=a_{1}+2\times 3\Rightarrow a_{1}=-1\Rightarrow S_{10}=\dfrac {2a_{1}+\left( n-1\right) d}{2}\times n=\dfrac {2\times \left( -1\right) +\left( 10-1\right) \times 3}{2}\times 10=125$
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