Solve the following


Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn − k Sn − 1 + Sn − 2 , then k =

(a) 1

(b) 2

(c) 3

(d) none of these


(b) 2

Let the A.P. be a, a+d, a+2d, a+3d...


$d=S_{n}-k S_{n-1}+S_{n-2}$

For n = 3, we have:

$d=(3 a+3 d)-k(2 a+d)+a$

$\Rightarrow 4 a+2 d-k(2 a+d)=0$

$\Rightarrow 2(2 a+d)=k(2 a+d)$

$\Rightarrow 2=k$


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