How do you prove the formula for the sum of n terms of an arithmetic progression?

An arithematic progression has terms in a sequence such that the difference between any two consecutive terms is a constant and is given by a common difference, say, d. If first term in the progression is a, then the second term would be a+d (i.e., the first term plus the common difference), and the third term would be a+d+d=a+2d, so on.. Therefore, following by that logic, the nth term of the series would be a+(n-1)d. To find the sum of the first n terms, we would have to add up all of these terms to find the sum as,

S = a + (a+d) + (a+2d) + .... + (a+(n-1)d)

If we club together all of the a's and all of the terms with the d's, we know that in n terms, there would a total of n a's giving a total sum of a's which is na (n multiplied by a)

And adding up all the d's would look something like: d+2d+...+(n-1)d = (1+2+...+(n-1))d which looks similar to the sum of the first (n-1) natural numbers which has a formula n(n-1)/2

Therefore, 

S = na + n(n-1)d/2 = (n/2)(a+(n-1)d)