(1) Sn = a1+A2+... an-1+an can also be written.

Sn=an+an- 1+......a2+a 1

The sum of the two formulas gives 2sn = (a1+an)+(a2+an-1)+... (an+a1).

=n(a 1+an)

So Sn=[n(a 1+an)]/2 (formula 1)

(2) If the first term of arithmetic progression is known as a 1, the tolerance is d and the number of terms is n, then an=a 1+(n- 1)d is substituted into the formula 1.

Sn = na1+[n (n+1) d]/2 (formula 2)