The 4^th  term of the sequence 6,4,3,… are in H.P is

Here $T_2-T_1=T_3-T_2=\frac{1}{12}$

$\Rightarrow \frac{1}{6},\frac{1}{4},\frac{1}{3}$  is an A.P

4^th term of this A.P. $=\frac{1}{6}+3\times \frac{1}{12}=\frac{1}{6}+\frac{1}{4}=\frac{5}{12}$

And the 4^th term  of H.P$=\frac{12}{5}$