It could be any number you like.
Given any fifth number it is easy to find a quartic polynomial (of degree 4) such that is passes through the given four points and the new one. Each choice of the fifth number will result in a different polynomial. So, since there are infinitely many choices for the fifth number, there are infinitely many position to value rules. In addition, there are non-polynomial functions as well.
The simplest answer is, perhaps, the cubic, Un = n3 - 9n2 + 32n - 10 for n = 1, 2, 3, ...
The sequence given is -2, -8, -18, -32, -50. To find the nth term, we first observe the differences between consecutive terms: -6, -10, -14, -18, which show that the second differences are constant at -4. This indicates that the nth term can be expressed as a quadratic function. By fitting the sequence to the form ( a_n = An^2 + Bn + C ), we find that the nth term is ( a_n = -2n^2 + 2n - 2 ).
If you mean: 2 4 8 16 32 64 it is 2^nth term and so the next number is 128
To find the nth term of the sequence -2, -8, -18, -32, -50, we first observe the differences between consecutive terms: -6, -10, -14, -18. The second differences (which are constant at -4) suggest that the nth term can be represented by a quadratic function. The general form is ( a_n = An^2 + Bn + C ). Solving for coefficients A, B, and C using the first few terms gives the nth term as ( a_n = -2n^2 + n ).
2n
37
[ 6n + 8 ] is.
8 + (74 x 6) = 75th term. nth term = 8 + 6(n-1)
It is: 9n+5 and so the next term is 50
The nth term is (36 - 4n)
5+9n
The sequence given is -2, -8, -18, -32, -50. To find the nth term, we first observe the differences between consecutive terms: -6, -10, -14, -18, which show that the second differences are constant at -4. This indicates that the nth term can be expressed as a quadratic function. By fitting the sequence to the form ( a_n = An^2 + Bn + C ), we find that the nth term is ( a_n = -2n^2 + 2n - 2 ).
If you meant: 2 12 22 32 then the nth term = 10n-8
If the nth term is 8 -2n then the 1st four terms are 6, 4, 2, 0 and -32 is the 20th term number
If you mean: 2 4 8 16 32 64 it is 2^nth term and so the next number is 128
To find the nth term of the sequence -2, -8, -18, -32, -50, we first observe the differences between consecutive terms: -6, -10, -14, -18. The second differences (which are constant at -4) suggest that the nth term can be represented by a quadratic function. The general form is ( a_n = An^2 + Bn + C ). Solving for coefficients A, B, and C using the first few terms gives the nth term as ( a_n = -2n^2 + n ).
2n
37