College Algebra (10th Edition)

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

Chapter 9 - Section 9.3 - Geometric Sequences; Geometric Series - 9.3 Assess Your Understanding - Page 664: 25


$a_n = \sqrt2 \cdot (\sqrt2)^{n-1}$ (This is equivalent to $a_n=(\sqrt{2})^{n}$.) $a_5=4\sqrt2$

Work Step by Step

RECALL: The $n^{th}$ term $a_n$ of a geometric sequence is given by the formula: $a_n=a_1 \cdot r^{n-1}$ where $a_1$ = first term $r$ = common ratio The given geometric sequence has: $a_1=\sqrt2$ $r=\sqrt2$ Thus, the $n^{th}$ term of the sequence is given by the formula: $a_n = \sqrt2 \cdot (\sqrt2)^{n-1}$ (This is equivalent to $a_n=(\sqrt{2})^{n}$.) The 5th term can be found by substituting $5$ for $n$: $a_5=\sqrt2 \cdot \sqrt2)^{5-1} \\a_5=\sqrt2 \cdot (\sqrt2)^4 \\a_5 = \sqrt2 \cdot 4 \\a_5=4\sqrt2$
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