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: 21


$a_1=10+\sqrt{7}$ $a_2=10$ $a_3=10-\sqrt{7}$ $a_4=10-2\sqrt{7}$ $a_5=10-3\sqrt{7}$

Work Step by Step

$a_1=10+\sqrt{7}$ $a_2=10$ Determine the common difference $d$: $d=a_2-a_1=10-(10+\sqrt{7})=-\sqrt{7}$ Then we have: $a_3=a_2+d=10+(-\sqrt{7})=10-\sqrt{7}$ $a_4=a_3+d=10-\sqrt{7}+(-\sqrt{7})=10-2\sqrt{7}$ $a_5=a_4+d=10-2\sqrt{7}+(-\sqrt{7})=10-3\sqrt{7}$
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