Q. 15

# Mark the correct

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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