Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - Chapter 11 Test Prep - Review Exercises - Page 1081: 9

Answer

$$3\pi - 2,2\pi - 1,\pi ,1, - \pi + 2$$

Work Step by Step

$$\eqalign{ & {\text{arithmetic;}}\,\,\,{a_3} = \pi ,\,\,{a_4} = 1 \cr & {\text{Calculate }}d \cr & d = {a_{n + 1}} - {a_n} \cr & d = {a_4} - {a_3} \cr & d = 1 - \pi \cr & \cr & {\text{The }}n{\text{th Term of an Arithmetic Sequence is}} \cr & {a_n} = {a_1} + \left( {n - 1} \right)d \cr & {a_n} = {a_1} + \left( {n - 1} \right)\left( {1 - \pi } \right) \cr & {\text{Let }}n = 3 \cr & {a_3} = {a_1} + \left( {3 - 1} \right)\left( {1 - \pi } \right) \cr & \pi = {a_1} + 2\left( {1 - \pi } \right) \cr & \pi = {a_1} + 2 - 2\pi \cr & {a_1} = 3\pi - 2 \cr & \cr & {\text{Then,}} \cr & {a_n} = 3\pi - 2 + \left( {n - 1} \right)\left( {1 - \pi } \right) \cr & \cr & {\text{Find }}{a_2},\,\,{a_5} \cr & {a_2} = 3\pi - 2 + \left( {2 - 1} \right)\left( {1 - \pi } \right) = 2\pi - 1 \cr & {a_5} = 3\pi - 2 + \left( {5 - 1} \right)\left( {1 - \pi } \right) = - \pi + 2 \cr & \cr & {\text{The terms are:}} \cr & 3\pi - 2,2\pi - 1,\pi ,1, - \pi + 2 \cr} $$
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