Arithmetic progression formula

Question

To find out the sum of arithmetic progression it has been given S=n(a+l)/2 And this formula has been derived in this manner $$ S=a+(a+d)+(a+2d)+.........+(l-2d)+(l-d)+l $$ And by reversing $$ S=l+(l-d)+(l-2d)+........+(a+2d)+(a+d)+a $$ On adding both we get $$ 2S=(a+l)+(a+l)+(a+l)+..........\text{ up to $n$ terms} $$ $$ 2S=n(a+l) $$ $$ S=n(a+l)/2 $$ My question is that why in deriving this formula reversing of the sum takes place(that is) $$ S=l+(l-d)+(l-2d)+........+(a+2d)+(a+d)+a $$ Why not any other method?