Answer
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:
$a_n=\frac{3}{2} + \frac{1}{2}(n-1)$
Work Step by Step
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:
$a_n=\frac{3}{2} + \frac{1}{2}(n-1)$
