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: 10

Answer

$${S_{10}} = 275$$

Work Step by Step

$$\eqalign{ & {S_{10}}{\text{ for the sequence 5,10,15,20,}}... \cr & {\text{Find }}d \cr & d = {a_{n + 1}} - {a_n} \cr & d = {a_2} - {a_1} \cr & d = 10 - 5 \cr & d = 5 \cr & \cr & {\text{The }}n{\text{th Term of an Arithmetic Sequence is given by}} \cr & {a_n} = {a_1} + \left( {n - 1} \right)d \cr & {\text{Then}} \cr & {a_n} = 5 + \left( {n - 1} \right)\left( 5 \right) \cr & {a_n} = 5 + 5n - 5 \cr & {a_n} = 5n \cr & \cr & {\text{The first term is}} \cr & {a_1} = 5 \cr & {\text{The last term is}} \cr & {a_{10}} = 5\left( {10} \right) = 50 \cr & {\text{Using the formula }}{S_n} = \frac{n}{2}\left( {{a_1} + {a_n}} \right),{\text{ we obtain}} \cr & {S_{10}} = \frac{{10}}{2}\left( {5 + 50} \right) \cr & {S_{10}} = 275 \cr} $$
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