The $n^{\text{th}}$  term of the H.P whose first two terms are 5 and 2 respectively

Second term of H.P is $\frac{1}{a+d}=2\Rightarrow a+d=\frac{1}{2}$

$\Rightarrow d=\frac{1}{2}-\frac{1}{5}=\frac{3}{10}$

$n^{\text{th}}$  term of A.P is $a+\left(n-1\right)d=\frac{1}{5}+\left(n-1\right)\frac{3}{10}=\frac{1}{5}+\frac{3n}{10}-\frac{3}{10}$

$\therefore n^{\text{th}}$  term of given H.P is $\frac{10}{3n-1}$