# Let Sn

Let a be the first term, n be the number of terms and d be the common difference of AP.

Given d = Sn – kSn–1 + Sn–2.

Now let n = 3

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

And d = S3 – k S3–1 + S3–2

d = S3 – k S2 + S1 ..............(1)

Sum of n terms of an AP is given as:

Sn = () x {2a + (n–1) d}

Now S1 = a

S2 = () x (2a + (2–1) d) (n =2)

S2 = (2a + d)

S3 = () x (2a + (3–1)d) (n =3)

S3 = x (2a + 2d)

S3 = 3(a + d)

S3 = 3a + 3d

Putting values of S1, S2 and S3 in equation 1, we get

d = 3a + 3d – k (2a + d) + a

d = 4a + 3d – k (2a + d)

k (2a + d) = 4a + 3d – d

k (2a + d) = 4a + 2d

k (2a + d) = 2(2a + d)

k = 2

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