College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.2 - Arithmetic Sequences - 9.2 Assess Your Understanding: 23

Answer

$a_n=\sqrt2+(n-1)(\sqrt2)$ $a_{51}=51\sqrt2$

Work Step by Step

RECALL: The $n^{th}$ term of an arithmetic sequence can be found using the formula: $a_n = a_1 + (n-1)d$ where $a_1$ = first term and $d$ = common difference The given sequence has: $a_1=\sqrt2$; $d=\sqrt2$ Substitute these values into the formula for the $n^{th}$ term to obtain: $a_n=a_1 + (n-1)d \\a_n=\sqrt2+(n-1)(\sqrt2)$ To find the 51st term, substitute $51$ for $n$ to obtain: $a_{51} = \sqrt2+(51-1)\sqrt2 \\a_{51}=\sqrt2+50\sqrt2 \\a_{51}=51\sqrt2$
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