If Sn denote the sum of n terms of an A.P. with first term a and common difference d such that S_xS_kx is independent of x, then

Answer: Option B

S_n is the sum of first n terms a is the first term and d is the common difference
$$\begin{array}{rl}& S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\\ & \frac{S_x}{S_{kx}}=\frac{\frac{n}{2}\left[2a+\left(n-1\right)d\right]}{\frac{kx}{2}\left[2a+\left(kx-1\right)d\right]}\\ & \because \frac{S_x}{S_{kx}}\text{is}\text{independent of}x\end{array}$$
$$\therefore \frac{\frac{n}{2}\left[2a+\left(x-1\right)d\right]}{\frac{kx}{2}\left[2a+\left(kx-1\right)d\right]}$$    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