Let the terms of the A.P. be:

a1 = a

a2 = a + d

a3 = a + 2d

a4 = a + 3d

a5 = a + 4d

The sum of the 5 consecutive terms is:

S5 = (5/2)(a1 + a5)

and also S5 = 45:

45 = (5/2)(a + a + 4d)

(5/2)(2a + 4d) = 45

2a + 4d = 18 divide both sides by 2

a + 2d = 9

a = 9 - 2d

The sum of cube of 2nd and 4th term is 1944:

a2^3 + a4^3 = 1944

(a + d)^3 + (a + 3d)^3 = 1944

plug a = 9 - 2d into the above equation:

(9 - 2d + d)^3 + (9 - 2d + 3d)^3 = 1944

by solving we find :

d1 = 3

d2 = -3

For d1 = 3 we have a = 9 - 2*3 = 3 thus the terms are:

3, 6, 9, 12 and 15

For d2 = -3 we have a = 9 - 2*(-3) = 15 thus the terms are:

15, 12, 9, 6  and 3.