College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.2 - Arithmetic Sequences - 8.2 Exercises: 40


$d=\frac{1}{2}$ $a_5=\frac{19}{6}$ The $n^{th}$ term is given by: $a_n = \frac{7}{6}+\frac{1}{2}(n-1)$ $a_{100}=\frac{152}{3}$

Work Step by Step

The sequence is arithmetic so the terms have a common difference. The common difference $d$ can be found by subtracting any term to the next term in the sequence. Thus, $d=\frac{5}{3} - \frac{7}{6} \\d=\frac{10}{6}-\frac{7}{6} \\d=\frac{10-7}{6} \\d=\frac{3}{6} \\d=\frac{1}{2}$ The fifth term $a_5$ can be found by adding the common difference $\frac{1}{2}$ to the fourth term. The fourth term of the sequence is $\frac{8}{3}$. Thus, $a_5 = \frac{8}{3}+\frac{1}{2} \\a_5=\frac{16}{6} + \frac{3}{6} \\a_5=\frac{16+3}{6} \\a_5=\frac{19}{6}$ The $n^{th}$ term $a_n$ of an arithmetic sequence is given by the formula $a_n = a+d(n-1)$ where $a$ = first term and $d$ = common difference. The sequence has $a=\frac{7}{6}$ and $d=\frac{1}{2}$. Thus, the $n^{th}$ term is given by: $a_n = \frac{7}{6}+\frac{1}{2}(n-1)$ Substituting 100 to $n$ gives the 100th term as: $\require{cancel} a_{100} = \frac{7}{6}+\frac{1}{2}(100-1) \\a_{100} = \frac{7}{6}+\frac{1}{2}(99) \\a_{100} = \frac{7}{6}+\frac{99}{2} \\a_{100} = \frac{7}{6} + \frac{99\cdot3}{2\cdot3} \\a_{100} = \frac{7}{6} + \frac{297}{6} \\a_{100} = \frac{7+297}{6} \\a_{100} = \frac{304}{6} \\a_{100} = \frac{\cancel{2}(152)}{\cancel{2}(3)} \\a_{100}=\frac{152}{3}$
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