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 - Page 1024: 8

Answer

$${a_n} = 3n - 7$$

Work Step by Step

$$\eqalign{ & {\text{From the figure we can see that }}{a_1} = - 4,\,\,\,\,{a_2} = - 1 \cr & {\text{Find }}d \cr & d = {a_{n + 1}} - {a_n} \cr & d = {a_2} - {a_1} \cr & d = \left( { - 1} \right) - \left( { - 4} \right) \cr & d = 3 \cr & \cr & {\text{The }}n{\text{th Term of an Arithmetic Sequence is given by}} \cr & {a_n} = {a_1} + \left( {n - 1} \right)d \cr & {\text{Then}} \cr & {a_n} = - 4 + \left( {n - 1} \right)\left( 3 \right) \cr & {a_n} = - 4 + 3n - 3 \cr & {a_n} = 3n - 7 \cr} $$
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