Explanation

(b)
Let $S_{1} = 5 + 9 + 13 + ... + n$ terms
$\Rightarrow \ \ \ S_{1} = \frac{n}{2}\left\lbrack 2 \times 5 + (n - 1)4 \right\rbrack = n(3 + 2n)$
and $S_{2} = 7 + 9 + 11 + ... + 12$ terms
$= \frac{12}{1}\left\lbrack 2 \times 7 + (12 - 1)2 \right\rbrack = 6(36)$
$= 216$
Since, $\frac{S_{1}}{S_{2}} = \frac{5}{12}$ (given)
$\Rightarrow \frac{n(3 + 2n)}{216} = \frac{5}{12}$
$\Rightarrow 2n^{2} + 3n - 90 = 0$
$\Rightarrow (2n + 15)(n - 6) = 0$
$\Rightarrow n = 6$ $(\because n$ cannot be negative)