Question:
Question 8: If Sn denote the sum of n terms of an A.P. with first term a and common difference d such that Sx/Skx is independent of x , then
Answer:
Given that a is the first term and d is the common difference of the AP
So Sx = (x/2)*{2a + (x-1)d}
Again
Skx = (kx/2){2a + (kx-1)d}
Now
Sx /Skx = [(x/2)*{2a + (x-1)d}]/[(kx/2){2a + (kx-1)d}]
= [{2a + (x-1)d}]/[k{2a + (kx-1)d}]
= [2a + xd - d]/[k{2a + kxd - d}]
= [(2a - d) + xd]/[k{(2a - d) + kxd}]
If we put d = 2a then this term is independent of x
Now Sx /Skx is independent of x if d = 2a.
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