Question 104572
The general series is S=a+(a+d)+(a+2d)+(a+3d) +....+(l-d))+l where a is the first term,d is the common difference and l is the last term.

If we write this in reverse we get S=l+(l-d)+(l-2d)+(l-3d)+....+(a+d)+a.

So we have,
S=a+(a+d)+(a+2d)+(a+3d)+....+(l-d)+l (i)
and
S=l+(l-d)+(l-2d)+(l-3d)+....+(a+d)+a (ii)

Now add (i) and (ii) together, making sure that you add corresponding terms together and you get
2S=(a+l)+(a+l)+(a+l)....+(a+l)+(a+l).
And so
 2S= n(a+l)   (because there are n lots of (a+l))
So
S= [n(a+l)]/2 (iii)

But your last term or nth term can be written as a+(n-1)d so
l=a+(n-1)d
Now replace l in (iii) and we get
                       S=[n(a+a+(n-1)d)]/2
This simplifies to 

                       S=[n(2a+(n-1)d)]/2 Q.E.D.

(N.B. l is the letter L and 1 is the number ONE)