## College Algebra 7th Edition

The first five terms are: $a_1 = \frac{3}{2}$ $a_2 = 2$ $a_3 = \frac{5}{2}$ $a_4 = 3$ $a_5 = \frac{7}{2}$ The given sequence is arithmetic with $d=\frac{1}{2}$. The $n^{th}$ term is: $a_n=\frac{3}{2} + \frac{1}{2}(n-1)$
Find the first five terms by substituting 1, 2, 3, 4 and 5 to $n$ in the given formula. $a_1 = 1+\frac{1}{2}=\frac{2}{2} + \frac{1}{2}=\frac{3}{2}$ $a_2 = 1+\frac{2}{2}=1+1 = 2$ $a_3 = 1+\frac{3}{2} = \frac{2}{2}+\frac{3}{2}=\frac{5}{2}$ $a_4 = 1+\dfrac{4}{2} = 1+2=3$ $a_5 = 1+\frac{5}{2} = \frac{2}{2}+\frac{5}{2}=\frac{7}{2}$ A sequence is arithmetic if there exists common difference among consecutive terms. Note that: $2 - \frac{3}{2} = \frac{4}{2} -\frac{3}{2} = \frac{1}{2} \\\frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}$ Since consecutive terms have a constant (or common) difference of one-half, then the given sequence is arithmetic with $d=\frac{1}{2}$. The $n^{th}$ term $a_n$ of the sequence whose $a=\frac{3}{2}$ and $d=\frac{1}{2}$ is given by the formula: $a_n=\frac{3}{2} + \frac{1}{2}(n-1)$