Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.2 - Arithmetic Sequences - Exercise Set: 32

Answer

$a_n=-\frac{1}{4} + \frac{1}{4}(n-1)$; $a_{20} = \frac{9}{2}$

Work Step by Step

RECALL: (1) The $n^{th}$ term, $a_n$, of an arithmetic sequence can be found using the formula: $a_n=a_1 + d(n-1)$ where $a_1$ = first term $n$ = term number $d$ = common difference (2) The common difference $d$ can be determined using the formula $d= a_n-a_{n-1}$ where $a_n$ = $n^{th}$ term $a_{n-1}$ = term before $a_n$ The given arithmetic sequence has $a_1=-\frac{1}{4}$ and $d=\frac{1}{4}$. Substitute these values into the formula in (1) above to obtain the arithmetic sequence's formula for the general term: $a_n=-\frac{1}{4}+(\frac{1}{4})(n-1)$ Solve for the 20th term for the sequence using the formula above to obtain: $a_{20} = -\frac{1}{4} + \frac{1}{4}(20-1) \\a_{20} = -\frac{1}{4}+\frac{1}{4}(19) \\a_{20} = -\frac{1}{4}+\frac{19}{4} \\a_{20} = \frac{18}{4} \\a_{20} = \frac{9}{2}$
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