Any number that you choose can be the nth 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 in the nth position. 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, U(n) = (3*n^4 - 30*n^3 + 105*n^2 - 126*n - 120)/4 gives the next number as 0orU(n) = (19*n^4 - 190*n^3 + 665*n^2 - 806*n - 696)/24 gives the next number as 1and so on.
The simplest solution, though, based on a polynomial of order 3 is U(n) = 6*(x - 8)
6n+10
When n=30, 3n-1 = 89 .
There are infinitely many possible answers. But the simplest is Un = 33 - 3n for n = 1, 2, 3, ...
t(n) = 6 + 8n
fsedaz sd
90
There are not enough numbers to be certain. The rule for the nth term could be Subtract 6 from the previous term giving 30, 24, 18, etc or Multiply previous term by 0.8 giving 30, 24, 21.6, etc etc
6n+10
Clearly here the nth term isn't n25.
When n=30, 3n-1 = 89 .
You can see that all the numbers go up by 7. This means that the first part of the nth term rule for this sequence is 7n. Now, you have to find out how to get from 7 to 3, 14 to 10, 21 to 17 ... this is because we are going up in the 7 times table. To get from the seventh times table to the sequence, you take away four. So the answer is : 7n-4
There are infinitely many possible answers. But the simplest is Un = 33 - 3n for n = 1, 2, 3, ...
The nth term of the sequence is 3n-8 and so the 30th term is 3*30 -8 = 82
n'th term: n^2 + 5
The nth term of the sequence is given by 30 - 2n. To find the first 5 terms, we substitute n = 1, 2, 3, 4, and 5 into the formula. When n = 1, the first term is 30 - 2(1) = 28. When n = 2, the second term is 30 - 2(2) = 26. When n = 3, the third term is 30 - 2(3) = 24. When n = 4, the fourth term is 30 - 2(4) = 22. When n = 5, the fifth term is 30 - 2(5) = 20.
t(n) = 6 + 8n
fsedaz sd