The nth term of $H.P$ , whose first two terms are 6 and 3 respectively is

$a_1=\frac{1}{6},a_2=\frac{1}{3}$  in $A.P$

$d=\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$

$a_n=a+\left(n-1\right)d$

$=\frac{1}{6}+\left(n-1\right)\frac{1}{6}$

$a_n=\frac{n}{6}$  in $A.P$

$a_n=\frac{6}{n}$  in $H.P$