maths~AP

let a - d, a , a + d be three consecutive terms
a - d + a + a + d = 24
3a = 24
a = 8
(a - d)(a)(a + d) = 312
(a^2 - d^2) a = 312
8(64 - d^2) = 312
64 - d^2 = 39
d^2 = 25
d = 5 or d = -5
two possible sets
3, 8, 13 or 13, 8 , 3

Let a be the first term, then the next two are a + d and a + 2d where d is the difference.
The sum = 24
a + a + d + a + 2d = 24
3a + 3d = 24
a + d = 8

The product = 312
(a)(a+d)(a+2d) = 312
(a^2 + ad)(a + 2d) = 312
a^3 + a^2d + 2a^2d + 2ad^2 = 312
a^3 + 3a^2d + 2ad^2 = 312

Since a+d = 8, d=8-a
Thus,
a^3 + 3a^2(8-a) + 2a(8-a)^2 = 312
a^3 + 24a^2 - 3a^3 + 2a(64 - 16a + a^2) = 312
-2a^3 + 24a^2 + 128a - 32a^2 + 2a^3 = 312
-8a^2 + 128a - 312 = 0
8a^2 - 128a + 312 = 0
a^2 - 16a + 39 = 0
(a-13)(a-3) = 0
a=13 or a=3

If a = 3, a+d = 8
d = 5
Therefore the sequence becomes 3, 8, 13
If a = 13, a + d = 8
d = -5
Therefore the sequence becomes 13, 8, 3