If we set n = 1 then we will get S1 = a
So substituting n = 1 we get a = S1 = 5(1)^2-3(1) = 5-3 = 2.
Also S2 = sum of first 2 terms
Hence S2 - S1 = second term
S2 = 5(2)^2 - 3(2) = 5(4) -6 = 20 -6 = 14
S2-S1 = 14-2 = 12
So 1st term = 2, 2nd term = 12
Common difference = 12-2 = 10.
So The AP has a = 2 and d = 10