Explanation

(c)
Let $S_{n}$ and $S'$ be the sums of $n$ terms of two AP’s and $T_{11}$ and $T_{11}^{'}$ be the respective 11th term, then
$\frac{S_{n}}{S_{n}^{'}} = \ \frac{\frac{n}{2}\left\lbrack 2a + (n - 1)d \right\rbrack}{\frac{n}{2}\left( 2a^{'} + (n - 1)d^{'} \right\rbrack} = \frac{7n + 1}{4n + 27}$ (given)
$\Rightarrow \ \frac{a + \frac{(n - 1)}{2}d}{a^{'} + \frac{(n - 1)}{2}d'} = \frac{7n + 1}{4n + 27}$
Now put, $n = 21,$ we get
$\frac{a + 10d}{a^{'} + 10d'} = \frac{T_{11}}{T_{11}^{'}} = \frac{148}{111}$
$= \frac{4}{3}$