Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.2 - Arithmetic Sequences - 12.2 Exercises - Page 857: 38


$d=-3$ $a_{n}=14-3n$ $ a_{5}=14-3\times 5=-1$ $a_{100}=14-3\times 100=-286$

Work Step by Step

$d=a_{4+1}-a_{n}=8-11=5-8=2-5=-3$ $a_{n}=a+\left( n-1\right) d\Rightarrow a_{1}=a+\left( 1-1\right) d=a=11$ $a_{n}=a+\left( n-1\right) d=11+\left( n-1\right) \times \left( -3\right) =11-34+3=14-3n$ $a_{n}=14-3n\Rightarrow a_{5}=14-3\times 5=-1;a_{100}=14-3\times 100=-286$
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