Answer
$a_{n}=-\dfrac {1}{2}+\dfrac {1}{12}n$
$a_{1}=-\dfrac {5}{12}$
Work Step by Step
$a_{n}=a+\left( n-1\right) d;a_{m}=a+\left( m-1\right) d\Rightarrow a_{n}-a_{m}=\left( n-m\right) d$
$\Rightarrow a_{14}-a_{9}=\dfrac {2}{3}-\dfrac {1}{4}=\left( 14-9\right) \times d=\dfrac {5}{12}\Rightarrow d=\dfrac {\left( \dfrac {5}{12}\right) }{5}=\dfrac {1}{12}$
$a_{n}=a+\left( n-1\right) d\Rightarrow a_{9}=a+\left( 9-1\right) \times \dfrac {1}{12}=\dfrac {1}{4}\Rightarrow a=\dfrac {1}{4}-\dfrac {8}{12}=-\dfrac {5}{12}$
$a_{n}=a+\left( n-1\right) d=\dfrac {-5}{12}+\left( n-1\right) \dfrac {1}{12}=-\dfrac {5}{12}-\dfrac {1}{12}+\dfrac {1}{12}n=-\dfrac {6}{12}+\dfrac {1}{12}n=-\dfrac {1}{2}+\dfrac {1}{12}n$
$a_{n}=-\dfrac {1}{2}+\dfrac {1}{12}n\Rightarrow a_{1}=-\dfrac {1}{2}+\dfrac {1}{12}=\dfrac {-5}{12}$
