If S_n denote the sum of n terms of an A.P. with first term a and common difference d such that $$\frac{{{S_x}}}{{{S_{kx}}}}$$ is independent of x, then

Solution(By Examveda Team)

S_n 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