Mark the correct alternative in the following:
Let denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by
, then k =
We know Sn=Sn-2+an-1+an
=Sn-2+a+(n-2)d+a+(n-1)d
=Sn-2+2a+2nd-3d
Also, Sn-1=Sn-2+an-1
=Sn-2+a+(n-2)d
=Sn-2+a+nd-2d
Now, d=Sn-kSn-1+Sn-2
= Sn-2+2a+2nd-3d-k(Sn-2+a+nd-2d)+Sn-2
=(2-k)[Sn-2+a+nd]+d(2k-3)
Comparing coefficient of both the side we get,
2-k=0 and 2k-3=1
∴ k=2