Any number that you choose can be the nth number. It is easy to find a rule based on a polynomial of order 5 such that the first five numbers are as listed in the question followed by the chosen number as the nth. 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.
For example, if you want the 7th number to be 2, use
U(n) = (n^5 - 15*n^4 + 85*n^3 - 225*n^2 + 154*n + 120)/120 for n = 1, 2, 3, ...
If you want the 8th number to be 6, then use
U(n) = (n^5 - 15*n^4 + 85*n^3 - 225*n^2 -6*n + 440)/140 for n = 1, 2, 3 ...
The simplest solution, though, based on a polynomial of order to is
U(n) = 4 - 2*n
The nth term is (2n - 12).
If you mean: 3, 4, 5, 6 and 7 then nth term = n+2
16 - 4nor4 (4 - n)
You can see that you add 10 to the previous term to get the next term. Term number 1 2 3 4 Term 4 14 24 34 You can also say: Term number 1 2 3 4 Term 0*10+4 1*10+4 2*10+4 3*10+4 So the nth term would be 10(n-1)+4 Or if you expand it, it's 10n-6
The nth term is: 3n-7 and so the next number will be 11
n - 1
If the nth term is 8 -2n then the 1st four terms are 6, 4, 2, 0 and -32 is the 20th term number
t(n) = n(n - 3)
The nth term is (2n - 12).
(n2+n+2) / 2, starting with n=0.
The Nth term in the series is [ 2N ] .
They are: nth term = 6n-4 and the 14th term is 80
If you mean: 3, 4, 5, 6 and 7 then nth term = n+2
11
16 - 4nor4 (4 - n)
If you mean: 2 4 8 16 32 64 it is 2^nth term and so the next number is 128
You can see that you add 10 to the previous term to get the next term. Term number 1 2 3 4 Term 4 14 24 34 You can also say: Term number 1 2 3 4 Term 0*10+4 1*10+4 2*10+4 3*10+4 So the nth term would be 10(n-1)+4 Or if you expand it, it's 10n-6