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

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