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: 27

Answer

it is an arithmetic sequence $d=7$ $a_{n}=11+7\left( n-1\right) $

Work Step by Step

İn order a sequence to be arithmetic : $a_{n+1}-a_{n}=a_{n}-a_{n-1}\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$ So: $2\times \left( 4+7n\right) =\left( 4+7\left( n+1\right) \right) +\left( 4+7\left( n-1\right) \right) \Rightarrow 8+14n=4+7n+7+4+7n-7\Rightarrow 8+14n=8+14n$ So this is arithmetic sequence $a_{n}=a+\left( n-1\right) d=a-d+nd=4+7n\Rightarrow d=7;a-d=4\Rightarrow a=11$ $a_{n}=11+7\left( n-1\right) $
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