Without further restrictions it can be any of an infinite number of formulae.
For example, U{n} = (11n⁴ - 110x³ + 385x² - 518x + 176)/8 which gives the next term as 42.
However, if it is an ARITHMETIC SEQUENCE (as I suspect your teacher wants), then the nth term is found:
Common difference: (-3) - (-7) = 4
→ 0th term is -7 - 4 = -11
→ nth term U{n} = 4n - 11
The nth term is: 5-2n
It is: nth term = 5-4n and so the next term will be -19
If you mean -1 3 7 11 15 then the nth term is 4n-5 and so the next term will be 19
The sequence 1, 3, 5, 7, 9 is an arithmetic sequence where each term increases by 2. The nth term can be expressed as ( a_n = 2n - 1 ). Therefore, for any positive integer ( n ), the nth term of the sequence is ( 2n - 1 ).
The sequence 5, 8, 11, 14, 17 is an arithmetic progression where each term increases by 3. The first term (a) is 5, and the common difference (d) is 3. The nth term can be expressed using the formula: ( a_n = a + (n-1)d ). Therefore, the nth term is ( a_n = 5 + (n-1) \cdot 3 = 3n + 2 ).
The nth term is: 5-2n
The nth term of the sequence is 2n + 1.
It is: nth term = 5-4n and so the next term will be -19
If you mean -1 3 7 11 15 then the nth term is 4n-5 and so the next term will be 19
The sequence 1, 3, 5, 7, 9 is an arithmetic sequence where each term increases by 2. The nth term can be expressed as ( a_n = 2n - 1 ). Therefore, for any positive integer ( n ), the nth term of the sequence is ( 2n - 1 ).
5 first terms in n²+3
Un = 4n - 9
The nth term is 4n - 3
12 - 5(n-1)
2n+3. If 5 is the first term, then it is 2n + 3 (2×1 = 2 + 3 = 5 and 2×2 + 3 = 7)
Un = (-1)n*(2n - 1)
This is an arithmetic sequence with initial term a = 3 and common difference d = 2. Using the nth term formula for arithmetic sequences an = a + (n - 1)d we get an = 3 + (n - 1)(2) = 2n - 2 + 3 = 2n + 1.