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

Sum of n terms of an AP is given as: 

S_n = (n/2)*{2a + (n-1)d}

Now, S_n-1 = {(n-1)/2}*{2a + (n-1 -1)d}

=> S_n-1 = {(n-1)/2}*{2a + (n- 2)d}

and S_n-2 = {(n-2)/2}*{2a + (n-2 -1)d}

=> S_n-2 = {(n-2)/2}*{2a + (n- 3)d}

Now, S_n - 2S_n-1 + S_n-2

= (n/2)*{2a + (n -1)d} - 2{(n-1)/2}*{2a + (n -2)d} + {(n-2)/2}*{2a + ( n -3)d}

= (1/2){2a*n + n(n - 1)d - 4a(n - 1) - 2(n - 1)*(n -2)d + 2a(n - 2) + (n - 2)*(n - 3)d}

= (1/2){2a*n + n(n - 1)d - 4a*n + 4n - 2(n - 1)*(n -2)d + 2a*n - 4a + (n - 2)*(n - 3)d}

= (1/2){2a(n - 2n + 2 + n + 2) + d(n² - n - 2n² + 6n - 4 + n² - 5n + 6)}

= (1/2){2a * 0 + 2d}    

= 2d/2

= d

So, the value of S_n - 2S_n-1 + S_n-2 is d