Explanation

(b)
Let $a_{1},\ a_{2},\ a_{3}\ and\ d_{1},\ d_{2},\ d_{3}$ are the first term and common difference of the three AP’s respectively.
We have, $a_{1} = \ a_{2} = \ a_{3} = 1\ \ and\ d_{1} = 1,\ d_{2} = 2,\ d_{3} = 3$
Therefore,
$S_{1} = \frac{n}{2}(n + 1)$ …(i)
$S_{2} = \frac{n}{2}(2n)$ …(ii)
$S_{3} = \frac{n}{2}(3n - 1)$ …(iii)
On adding Eqs. (i) and (iii), we get
$S_{1} + S_{3} = \frac{n}{2}\left\lbrack (n + 1) + (3n - 1) \right\rbrack$
$= 2\left\lbrack \frac{n}{2}(2n) \right\rbrack = 2S_{2}$
Hence, correct relation is $S_{1} + S_{3} = 2\ S_{2}$