If Sn denote the sum of n terms of an A.P. with first term a and common difference dsuch that Sx/Skx is independent of x, then

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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 Sn. We need to find the relation between a and d such thatS_x/S_(kx) is independent of So, let us first find the values of Sx and Skx 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

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