Any number that you choose can be the 15th number. It is easy to find a rule based on a polynomial of order 4 such that the first four numbers are as listed in the question followed by the chosen number as the 15th term. There are also non-polynomial solutions. 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.
One possible solution is to use the following polynomial of order 3 for the first 4 numbers:
U(n) = (-32*n^3 + 228*n^2 - 478*n + 291)/3 for n = 1, 2, 3, ...
so that U(15) = -21193
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every next term is 4 smaller than previous so 7th term = -23
after -9 it is -15 then -21, -27 and the ninth is -36
Double it minus the previous number.
The nth term in the sequence -5, -7, -9, -11, -13 can be represented by the formula a_n = -2n - 3, where n is the position of the term in the sequence. In this case, the common difference between each term is -2, indicating a linear sequence. By substituting the position n into the formula, you can find the value of the nth term in the sequence.
To find the common ration in a geometric sequence, divide one term by its preceding term: r = -18 ÷ 6 = -3 r = 54 ÷ -18 = -3 r = -162 ÷ 54 = -3