Let Sn denote the sum of n terms of an A.P. whose first term is a.

In the given problem, we are given $d=S_{n}-k S_{n-1}+S_{n-2}$

We need to find the value of k

So here,

First term = a

Common difference = d

Sum of n terms = Sn

Now, as we know,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$ .........(1)

Also, for n-1 terms,

$S_{n-1}=\frac{n-1}{2}[2 a+[(n-1)-1] d]$

$=\frac{n-1}{2}[2 a+(n-2) d]$........(2)

Further, for n-2 terms,

$S_{n-2}=\frac{n-2}{2}[2 a+[(n-2)-1] d]$

$=\frac{n-2}{2}[2 a+(n-3) d]$ ...........(3)

Now, we are given,

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

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

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

Using (1), (2) and (3) in the given equation, we get

$k=\frac{\frac{n}{2}[2 a+(n-1) d]+\frac{n-2}{2}[2 a+(n-3) d]-d}{\frac{n-1}{2}[2 a+(n-2) d]}$

Taking $\frac{1}{2}$ common, we get,

$k=\frac{n[2 a+(n-1) d]+(n-2)[2 a+(n-3) d]-2 d}{(n-1)[2 a+(n-2) d]}$

$=\frac{2 a n+n^{2} d-n d+2 a n+n^{2} d-3 n d-4 a-2 n d+6 d-2 d}{2 a n+n^{2} d-2 n d-2 a-n d+2 d}$

$=\frac{2 n^{2} d+4 a n-6 n d-4 a+4 d}{n^{2} d+2 a n-3 n d+-2 a+2 d}$

Taking 2 common from the numerator, we get,

$k=\frac{2\left(n^{2} d+2 a n-3 n d+-2 a+2 d\right)}{n^{2} d+2 a n-3 n d+-2 a+2 d}$

$=2$

Therefore, $k=2$

Hence, the correct option is (b).