Answer :

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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

There are n A.M.sRD Sharma - Mathematics

If x, y, z are inRD Sharma - Mathematics

Insert 7 A.M.s beRD Sharma - Mathematics

The 10th</suRD Sharma - Mathematics

Insert five numbeRD Sharma - Mathematics

The 4th</supRD Sharma - Mathematics

Insert 4 A.M.s beRD Sharma - Mathematics

Show that x2RD Sharma - Mathematics

An A.P. consists RD Sharma - Mathematics

How many terms arRS Aggarwal - Mathematics