Question

If S_n denote the sum of n terms of an A.P. with first term a and common difference dsuch that \[\frac{Sx}{Skx}\]  is independent of x, then

 

Options

•  d= a

• d = 2a

• a = 2d

• d = −a

Answer 1

Here, we are given an A.P. with a as the first term and d as the common difference. The sum of n terms of the A.P. is given by S_n.

We need to find the relation between a and d such that`S_x/S_(kx)` is independent of

So, let us first find the values of S_x and S_kx using the following formula for the sum of n terms of an A.P.,

`S_n = n/2 [ 2a + ( n- 1) d ]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, we get,

`S_x = x/2 [ 2a + ( x - 1) d ]`

Similarly,

`S_(kx) = (kx)/2 [ 2a + ( kx - 1) d ]`

So,

`S_x /S_(kx) = (x/2[2a + (x -1)d ] )/((kx)/2 [2a + (kx - 1 ) d])`

        `=([2a + ( x -1) d ])/(k[2a + (kx -1) d ]) `

        `=(2a + dx - d)/(2ak + k^2 xd -kd)`

 Now, to get a term independent of x we have to eliminate the other terms, so we get

2a - d = 0

    2a  =  d 

So, if we substitute 2a = d , we get,

  `(2a + dx - d)/(2ak + k^2 xd -kd)=(2a + dx -2a)/(2ak + k^2 xd -2ak)`

                                  `=(dx)/(k^2 dx)`

                                  `= 1/(k^2)`

Therefore, 2a = d