It could be
Un = (-3n4 + 30n3 - 105n2 + 168n - 80)/2
or
Un = (-5n4 + 50n3 - 175n2 + 286n - 136)/4
or any one of infinitely more polynomials of order >1, or other types of functions.
The simplest polynomial solution is Un = 9n - 4
or any
It is: 9n+5 and so the next term is 50
5+9n
[ 6n + 8 ] is.
The nth term is (36 - 4n)
8 + (74 x 6) = 75th term. nth term = 8 + 6(n-1)
It is: 9n+5 and so the next term is 50
5+9n
If you meant: 2 12 22 32 then the nth term = 10n-8
[ 6n + 8 ] is.
The nth term is (36 - 4n)
8 + (74 x 6) = 75th term. nth term = 8 + 6(n-1)
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 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
2n
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 ).
37