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: 40


$d=\frac{1}{2}$ $a_{n}=\dfrac {2}{3}+\dfrac {1}{2}n $ $ a_{5}=\dfrac {19}{6};$ $a_{100}=\dfrac {152}{3}$

Work Step by Step

$d=a_{n+1}-a_{n}=\dfrac {5}{3}-\dfrac {7}{6}=\dfrac {13}{6}-\dfrac {5}{3}=\dfrac {8}{3}-\dfrac {13}{6}=\dfrac {3}{6}=\dfrac {1}{2}$ $a_{n}=a+\left( n-1\right) d\Rightarrow a_{1}=a+\left( 1-1\right) d=a=\dfrac {7}{6}$ $a_{n}=a+\left( n-1\right) d=\dfrac {7}{6}+\left( n-1\right) \dfrac {1}{2}=\dfrac {7}{6}-\dfrac {1}{2}+\dfrac {1}{2}n=\dfrac {7-3}{6}+\dfrac {1}{2}n=\dfrac {4}{6}+\dfrac {1}{2}n=\dfrac {2}{3}+\dfrac {1}{2}n$ $a_{n}=\dfrac {2}{3}+\dfrac {1}{2}n\Rightarrow a_{5}=\dfrac {2}{3}+\dfrac {1}{2}\times 5=\dfrac {19}{6};a_{100}=\dfrac {2}{3}+\dfrac {1}{2}\times 100=\dfrac {152}{3}$
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