How do I find the sum of a sequence whose common difference is in Arithmetic Progression? How do I find the sum of a sequence whose common difference is in Arithmetic Progression ?
Like in the following series :-
$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91$
And also how to find it's $n^{th}$ term ??

I enjoyed several of the solutions here. However the idea of substituting values and solving for a, b, c seems really long and inefficient.
Below I have an efficient method to do this. 
As codetalker pointed out:
a(n+1)-a(n)= some linear term=1+n(for example) 
So now what you do is make a telescopic series:
a(2)-a(1)=1+1
a(3)-a(2)=1+2
........................
a(n)-a(n-1)=1+(n-1)
So now you add all individual equations, you see that
a(n) =a(1)+n×1+(n-1)(n)/2 
( I have used the formula for sum of n terms).
You know a(1)
So now you have a(n) 
On summation you can apply the formulae the other answers specified for sum of numbers and squares.
Hope this answer makes is more efficient to do these
