Question

Let $S_1,S_2,S_3--------S_n$ denotes the sum of 1, 2, 3 ----n terms of an A.P, having first term a. If $\frac{S_{kx}}{S_x}$ k $\ne$ 1, is independent of x, then $S_1+S_2+S_3--------S_n$ =

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We will start with the given condition and try to get more information/relations connecting a and d.

$\frac{S_{kx}}{S_x}=\frac{\frac{kx}{2}[2a+(kx-1\left)d\right]}{\frac{x}{2}[2a+(x-1\left)d\right]}$

$=\frac{k\left[\right(2a-d)+kxd]}{\left[\right(2a-d)+xd]}$

For this to be independent of x, 2a-d=0

[Another way of finding the condition is finding the derivative with respect to x and equating it to zero]

$\Rightarrow$ d=2a

$\therefore S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}[2a+(n-1\left)2a\right]$

$=\frac{n}{2}$ x 2a x n

$=n^{2}a$

$\Rightarrow S_1,S_2,S_3----S_n=1^{2}a+2^{2}a---n^{2}a$

$=(1^2+2^2----n^2)a$

$=\frac{n(n+1)(2n+1)}{6}a$