Explanation

(b)
Let sum of $2n$ terms of the AP 2, 5, 8, 11, … is $S_{2n}$
$\therefore$ $S_{2n} = \frac{2n}{2}\lbrack 2 \times 2 + (2n - 1)3\rbrack$
$= n(4 + 6n - 3)$
$= n(6n + 1)$
And sum of $n$ terms of the AP 57, 59, 61, 63,… is $S_{n}$
$\therefore\ \ S_{n} = \frac{n}{2}\lbrack 2 \times 57 + (n - 1)2\rbrack$
$= \frac{n}{2}(2n + 112)$
According to question, $S_{2n} = S_{n}$
$\Rightarrow n(6n + 1) = \frac{n}{2}(2n + 112)$
$\Rightarrow 12n + 2 = 2n + 112$
$\Rightarrow 10n = 110$
$\Rightarrow n = 11$