if Sn denotes the sum if n terms of an A.P whose common difference is is d ,show that

d=S_n-2S_n-1+S_n-2

 
Sn = a + ( a + d ) + ( a + 2d ) + ..+ ( a + n - 3d) + (a + n - 2 d ) + ( a + n - 1 d)----(1) 

Rewriting Sn = (a + n -1 d) + ( a + n - 2 d) + (a + n - 3 d ) + .....+ (a + 2d) + ( a + d ) +a---(2)

Now we add (1) and (2) term by term . We observe that the sum of any term in (1) and the corresponding term in (2) is 2a + ( n - 1) . For instance how many times will get ( 2a = (n - 1)d) ? It is clear, that each Sn in (1) and (2) has n terms and therefore, we have 

2 Sn = n {2a + (n - 1) d }

or S n = n/2 { 2a + ( n-1 ) d }, the formula for finding sum of n terms of an A. P .