In a H.P. 4^th term is $\frac{1}{9}$ and 13^th term is $\frac{1}{27}$, then general term is 

Given $a_4=\frac{1}{a+3d}=\frac{1}{9}\Rightarrow a+3d=9\cdots \cdots \cdots \left(1\right)$

And $a_{13}=\frac{1}{a+12d}=\frac{1}{27}\Rightarrow a+12d=27\cdots \cdots \cdots \left(2\right)$

$\left(2\right)-\left(1\right)\Rightarrow 9d=18\Rightarrow d=\frac{18}{9}=2$

Substitute $d=2$ in equation (1)  $\Rightarrow a+3\left(2\right)=9\Rightarrow a=9-6=3$

$\therefore$ General term $a_n=\frac{1}{a+(n-1)d}=\frac{1}{3+(n-1)2}=\frac{1}{3+2n-2}=\frac{1}{2n+1}$