There are infinitely many polynomials of order 6 that will give these as the first six numbers and any one of these could be "the" rule. 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) = 0.1*n^6 - 2.1*n^5 + 17.5*n^4 - 73.5n^3 + 164.4*n^2 - 206.4*n + 184 will do the trick.
The simplest solution, based on a polynomial of degree 2, is
V(n) = 2*n^2 - 30*n + 112.
If you mean: 6 12 18 24 then the nth term is 6n
The nth term of that series is (24 - 6n).
If you mean: 34 39 24 ... then the nth term is 39-5n and so the 100th term = -461
The nth term of an AP with initial term a (= u{1}) and common difference d is given by: u{n} = a + (n - 1)d In this case: a = 6 d = (12 - 6) = 6 → u{n} = 6 + (n - 1)6 But this can be simplified: u{n} = 6 + (n - 1)6 = 6 + 6n - 6 = 6n
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
Give the simple formula for the nth term of the following arithmetic sequence. Your answer will be of the form an + b.12, 16, 20, 24, 28, ...
For {12, 15, 18} each term is the previous term plus 3; a general formula for the nth term is given by t(n) = 3n + 9. For {12, 24, 36} each term is the previous term plus 12; a general formula for the nth term is given by t(n) = 12n.
If 3 is the first term, then the nth term is [ 3 x 2(n-1) ] .
If 3 is the first term, then the nth term is [ 3 x 2(n-1) ] .
If you mean: 6 12 18 24 then the nth term is 6n
The nth term of that series is (24 - 6n).
44
24 - 6n
The nth term is given by: rn = n2 + 8
The nth term is (36 - 4n)
2n(n+1)
8 + 4n