I think that the wording of the question leaves much to be desired, but here's my interpretation of it.
As a preamble, the sum to n terms of the arithmetic progression with first term a and common difference d is (n/2)(2a + (n-1)d).
I take it that the first payment is a, the second a+d, the third a+2d and so on.
The intention was that 3600 was paid off in 40 instalments, so
3600 = (40/2)(2a + (40-1)d) , that is, 180 = 2a + 39d.
Two thirds are paid off after 30 instalments, so
2400 = (30/2)(2a + (30-1)d), or, 160 = 2a + 29d.
Solve those and you get a=51 and d=2.