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 − kSn−1 + Sn−2, then k =

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In the given problem, we are given d = Sn - kSn-1 + Sn-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 = n/2 [2a + (n - 1) d ] .............(1) Also, for n-1 terms, `S_(n-1) = (n-1)/2 [ 2a + [(n-1) - 1]d] = (n-1)/2 [2a + ( n -2) d ] ` .............(2) Further, for n-2 terms, `S_(n - 2) = (n - 2)/2 [ 2a + [(n - 2 ] d] = (n-2)/2 [ 2a + ( n - 3) d ] ` ...............(3) Now, we are given, d = Sn - kSn -1 + Sn-2 d + KSn-1 = Sn + Sn-2 ` K = (S_n + S_(n-2) - d ) /S_(n-1)` Using (1), (2) and (3) in the given equation, we get `k = (n/2[2a + (n-1) d ] + (n-2)/2 [2a + ( n - 3)d]-d)/((n-1)/2 [ 2a + (n - 2 ) d ]) Taking $\frac{1}{2}$ common, we get, K = (n[2a + (n-1)d] + (n-2) [2a + ( n- 3)d]- 2d)/((n-1)[2a + (n-2) d]) =(2an + n^2d - nd + 2an + n^2d - 3nd - 4a - 2nd + 6d - 2d)/(2an + n^2d - 2nd - 2a - nd + 2d) = (2n^2d + 4an - 6nd - 4a + 4d)/(n^2d + 2an - 3nd + -2a + 2d)` Taking 2 common from the numerator, we get, `k = (2(n^2 d + 2an - 3nd + -2a + 2d))/(n^2d + 2an - 3nd + -2a +2d)` = 2 Therefore, k = 2

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