#### Answer

$a=-\dfrac{5}{12}$
$a_n = -\dfrac{5}{12} + \dfrac{1}{4}(n-1)$

#### Work Step by Step

RECALL:
The $n^{th}$ $a_n$ of an arithmetic sequence is given by the formula:
$a_n = a + d(n-1)$
where
$a$ = first term
$d$ = common difference
The fourteenth term is $\frac{2}{3}$ . This means that:
$a_{14} = a + d(14-1)
\\\frac{2}{3} = a + d(13)
\\\frac{2}{3} = a +13d$
The ninth term is $\frac{1}{4}$. This means that:
$a_9=a+d(9-1)
\\\frac{1}{4}=a + d(8)
\\\frac{1}{4} = a + 8d$
The given information about the sequence results to two equations:
(equation 1) $\frac{2}{3}=a+13d$
(equation 2) $\frac{1}{4}=a+8d$
Subtract equation 2 to equation 1 to obtain:
$
\begin{array}{cccc}
& &\frac{2}{3} &= &a +13d
\\&-&&&&&
\\& &\frac{1}{4} &= &a + 8d
\\&&\text{_____}&\text{______}&\text{_________}
\\&&\frac{5}{12} &= &5d\end{array}$
Divide both sides by $5$ to obtain:
$\dfrac{\frac{5}{12}}{5}=\dfrac{5d}{5}
\\\frac{5}{12} \cdot \frac{1}{5} = d
\\\frac{1}{12}=d$
Solve for $a$ by substituting $d=\frac{1}{12}$ to Equation 2 to obtain:
$\require{cancel}
\frac{1}{4} = a + 8d
\\\frac{1}{4} = a + 8(\frac{1}{12})
\\\frac{1}{4} = a + \frac{8}{12}
\\\frac{1}{4} = a + \frac{2\cancel{(4)}}{3\cancel{(4)}}
\\\frac{1}{4} = a + \frac{2}{3}
\\\frac{1}{4} - \frac{2}{3} = a
\\\frac{3}{12} = \frac{8}{12} = a
\\\frac{3-8}{12}=a
\\\frac{-5}{12}=a$
Thus, the first term is $a=-\dfrac{5}{12}$.
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
$d$ = common difference
The given sequence has $a= -\dfrac{5}{12}$ and $d=\dfrac{1}{12}$.
Substituting these to the nth term formula gives:
$a_n = -\dfrac{5}{12} + \dfrac{1}{4}(n-1)$