Explanation

(a)
Since, $\frac{\left( S_{n} \right)_{1}}{\left( S_{n} \right)_{2}} = \frac{2n + 3}{6n + 5}$ …(i)
$\Rightarrow \ \frac{\frac{n}{2}\lbrack 2a_{1} + (n - 1)d_{1}\rbrack}{\frac{n}{2}\lbrack 2a_{2} + (n - 1)d_{2}\rbrack} = \frac{2n + 3}{6n + 5}$
$\Rightarrow \ \frac{a_{1} + \frac{(n - 1)}{2}d_{1}}{a_{2} + \frac{(n - 1)}{2}d_{2}} = \frac{2n + 3}{6n + 5}$
Put $\frac{n - 1}{2} = 12\ \Rightarrow n = 25$
$\therefore\ \frac{a_{1} + 12d_{1}}{a_{2} + 12d_{2}} = \frac{53}{155}$
$\Rightarrow \ \ \frac{\left( T_{13} \right)_{1}}{\left( T_{13} \right)_{2}} = \frac{53}{155}$