#### Answer

$d=\frac{1}{2}$
$a_5=\frac{19}{6}$
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:
$\require{cancel}
a_{100} = \frac{7}{6}+\frac{1}{2}(100-1)
\\a_{100} = \frac{7}{6}+\frac{1}{2}(99)
\\a_{100} = \frac{7}{6}+\frac{99}{2}
\\a_{100} = \frac{7}{6} + \frac{99\cdot3}{2\cdot3}
\\a_{100} = \frac{7}{6} + \frac{297}{6}
\\a_{100} = \frac{7+297}{6}
\\a_{100} = \frac{304}{6}
\\a_{100} = \frac{\cancel{2}(152)}{\cancel{2}(3)}
\\a_{100}=\frac{152}{3}$