If the 2nd, the 31st and the nth term of an Arithmetic progression are respectively 7\(\frac{3}{4}\) , \(\frac{1}{2}\) and -6\(\frac{1}{2}\) then the sum of the first n terms of this progression is
1
44\(\frac{1}{4}\)
2
48\(\frac{1}{2}\)
3
51\(\frac{1}{3}\)
4
52\(\frac{3}{4}\)