Answer: Option B
Sn is the sum of first n terms a is the first term and d is the common difference
$$\eqalign{
& {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr
& \frac{{{S_x}}}{{{S_{kx}}}} = \frac{{\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}{{\frac{{kx}}{2}\left[ {2a + \left( {kx - 1} \right)d} \right]}} \cr
& \because \frac{{{S_x}}}{{{S_{kx}}}}\,{\text{is}}\,{\text{independent of}}\,x \cr} $$
$$\therefore \frac{{\frac{n}{2}\left[ {2a + \left( {x - 1} \right)d} \right]}}{{\frac{{kx}}{2}\left[ {2a + \left( {kx - 1} \right)d} \right]}}\,$$    is independent of x
$$\therefore \frac{{\frac{n}{2}\left[ {2a + xd - d} \right]}}{{\frac{{kx}}{2}\left[ {2a + kdx - d} \right]}}$$    is independent of x
$$ \Rightarrow \frac{{2a - d}}{{k\left( {2a - d} \right)}}$$   is in dependent of x if 2a - d $$ \ne $$ 0
If 2a - d =0, then d = 2a