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: 47

Answer

$a_{n}=-\dfrac {1}{2}+\dfrac {1}{12}n$ $a_{1}=-\dfrac {5}{12}$

Work Step by Step

$a_{n}=a+\left( n-1\right) d;a_{m}=a+\left( m-1\right) d\Rightarrow a_{n}-a_{m}=\left( n-m\right) d$ $\Rightarrow a_{14}-a_{9}=\dfrac {2}{3}-\dfrac {1}{4}=\left( 14-9\right) \times d=\dfrac {5}{12}\Rightarrow d=\dfrac {\left( \dfrac {5}{12}\right) }{5}=\dfrac {1}{12}$ $a_{n}=a+\left( n-1\right) d\Rightarrow a_{9}=a+\left( 9-1\right) \times \dfrac {1}{12}=\dfrac {1}{4}\Rightarrow a=\dfrac {1}{4}-\dfrac {8}{12}=-\dfrac {5}{12}$ $a_{n}=a+\left( n-1\right) d=\dfrac {-5}{12}+\left( n-1\right) \dfrac {1}{12}=-\dfrac {5}{12}-\dfrac {1}{12}+\dfrac {1}{12}n=-\dfrac {6}{12}+\dfrac {1}{12}n=-\dfrac {1}{2}+\dfrac {1}{12}n$ $a_{n}=-\dfrac {1}{2}+\dfrac {1}{12}n\Rightarrow a_{1}=-\dfrac {1}{2}+\dfrac {1}{12}=\dfrac {-5}{12}$
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