Find the 4^th and 8^th term of the series 6, 4, 3, …

$\begin{array}{l}\frac{1}{6},\frac{1}{4},\frac{1}{3},\cdots \\ t_2-t_1=\frac{1}{4}-\frac{1}{6}=\frac{3-2}{12}=\frac{1}{12}\end{array}$

$\therefore \frac{1}{6},\frac{1}{4},\frac{1}{3},\cdots$ are in A.P

$\Rightarrow a=\frac{1}{6},d=\frac{1}{12},n=4$

4^th term $=a+(n-1)d$

$\begin{array}{l}=\frac{1}{6}+(4-1)\times \frac{1}{12}\\ \Rightarrow \frac{1}{6}+3\times \frac{1}{12}\\ =\frac{1}{6}+\frac{1}{4}=\frac{2+3}{12}\\ =\frac{5}{12}\end{array}$

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

8^th term $=a+7d$

$\begin{array}{l}=\frac{1}{6}+7\times \frac{1}{12}\\ =\frac{1}{6}+\frac{7}{12}=\frac{2+7}{12}=\frac{9}{12}\end{array}$

$\therefore$ 8^th term of H.P.$=\frac{4}{3}$