Question

let sn denote sum of n terms of an a p whose first term is a .if common difference is d=Sn-kSn-1+Sn-2,then find k

Answer 1

Sn-1 = Sn-2 + a_(n-1)
= Sn-2 + (a + (n-2)d)

Sn = Sn-2 + a_(n-1) + a_n
= Sn-2 + (a + (n-2)d) + (a + (n-1)d)

So

Sn - kSn-1 + Sn-2 = (2-k)Sn-2 + (a + (n-2)d) + (a + (n-1)d) - k(a + (n-2)d)
= (2-k)Sn-2 + (2-k)a + d[(2-k)n + 2k-3]
= (2-k)(Sn-2 + a + nd] + d(2k-3)

This is to equal d, which requires

2-k = 0

and

2k-3 = 1

It is fortunate that the same value of k satisfies both of these: k=2