Question

In an A.P., prove that $d=S_n-2S_{n-1}+S_{n-2}$.

Answer 1

For an Arithmetic Progression,

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

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

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

Now, let's take the $RHS$ :

$RHS=S_n-2S_{n-1}+S_{n-2}=an+\frac{n(n-1)}{2}.d-2a(n-1)-(n-1)(n-2)d+a(n-2)+\frac{(n-2)(n-3)}{2}.d=a[n-2(n-1)+n-2]+d[\frac{n(n-1)}{2}-(n-1\left)\right(n-2)+\frac{(n-2)(n-3)}{2}]=0+\frac{d}{2}\left[\right(n^2-n)-2(n^2-3n+2)+(n^2-5n+6\left)\right]=\frac{d}{2}(-4+6)=\frac{d}{2}\times 2=d=LHS$

$LHS=RHS$

Thus it is proved.