There are infinitely many polynomials of order 4 that will give these as the first four numbers and any one of these could be "the" rule. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one. Each one of these so-called rules will give a different set of the next three.
Conversely, you can select any three numbers to be the next numbers and there will be a rule based on a polynomial of order six that will fit the above four and your selection of three.
For example, you could select the rule as
U(n) = (-465*n^6 + 11334*n^5 - 108805*n^4 + 520570*n^3 - 1289870*n^2 + 1539956*n - 674160)/120 for n = 1, 2, 3, ...
The resulting sequence is -12, 24, -48, 96, 1, 2, 3, ...
Or, if the rule is
U(n) = (-228*n^6 + 5559*n^5 - 53390*n^4 + 255605*n^3 - 633802*n^2 + 757216*n - 331680)/60 for n = 1, 2, 3, ...
then resulting sequence is -12, 24, -48, 96, 10, 20, 30, ...
The simplest solution, based on
U(1) = -12
U(n+1) = -2*U(n) for n = 1, 2, 3, ...
or, equivalently by U(n) = -6*2^n
gives the following sequence: -12, 24, -48, 96, -192, 384, -768,
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The difference between the successive numbers is 4. so the next three numbers could be 4, 0 and -4.
12, 13, 16 Add three, then add 1, then add three, then add 1, and so on so forth
12, 10 and 15 the are alternatively increased by 2 and 3
11, -10, 9
-3 0 3 then 6 9 12 15 18 21 and so on ...