Explanation

(b)
Let the first term of an AP be $a$ and common difference be $d$.
Since, $a + 3d = \frac{5}{3}$ …(i)
and $a + 7d = 3$ …(ii)
On solving Eqs.(i) and (ii), we get
$a = \frac{2}{3}\ ,\ \ \ d = \frac{1}{3}$
$\therefore T_{6} = a + 5d = \frac{2}{3} + \frac{5}{3} = \frac{7}{3}$
$\Rightarrow$ 6th term of HP is $\frac{3}{7}$.