Let `S_n` denote the sum of n terms of an AP whose first term is a. If common difference d is given by `d=Sn-kS_(n-1)+S_(n-2)` , then k is :

Explanation

Given `d=Sn-kS_(n-1)+S_(n-2)`

Now let `n = 3`

So, AP is
: a, a + d, a + 2d

And `d = S_3 – k S_(3–1) + S_(3–2)`

=>` d = S_3 – k S_2 + S_1` ..............(1)

Sum of n terms of an AP is given as:

` S_n = (n/2 ) xx {2a + (n–1) d} `

Now `S_1 = a`

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