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

A. d = a B. d = 2a C. a = 2d D. d = -a Answer: Option B

### Solution(By Apex Team)

Sn is the sum of first n terms a is the first term and d is the common difference \begin{aligned}&S_n=\frac{n}{2}[2a+(n-1)d]\\ &\frac{S_x}{S_{kx}}=\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{kx}{2}[2a+(kx-1)d]}\\ &\because\frac{S_x}{S_{kx}}\text{ is independent of }x\\ &\therefore\frac{\frac{n}{2}[2a+(x-1)d]}{\frac{kx}{2}[2a+(kx-1)d]}\text{ is independent of }x\\ &\therefore\frac{\frac{n}{2}[2a+xd-d]}{\frac{kx}{2}[2a+kdx-d]}\text{ is independent of }x\\ &\Rightarrow\frac{2a-d}{k(2a-d)}\text{ is in dependent of }x\text{ if }2\mathrm{a}-\mathrm{d}\ne0\end{aligned} If 2a – d =0, then d = 2a

A. 22
B. 25
C. 23
D. 24

A. 5
B. 6
C. 4
D. 3

A. -45
B. -55
C. -50
D. 0