Given any 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 and the nth is the given number. There are also non-polynomial solutions.
The simplest polynomial of smaller order is:
Un = (n2 + n + 4)/2 for n = 1, 2, 3, ...
The nth term is 18 -3n and so the next term will be 3
t(n) = 5n - 3
3n
12 - 5(n-1)
If 3 is the first term, then the nth term is [ 3 x 2(n-1) ] .
The nth term is 5n-3 and so the next term will be 22
5n - 3
tn=5n-3
The nth term is 18 -3n and so the next term will be 3
t(n) = 5n - 3
The nth term in the arithmetic progression 10, 17, 25, 31, 38... will be equal to 7n + 3.
To find the nth term of the sequence 9, 12, 17, 24, 33, we first look at the differences between consecutive terms: 3, 5, 7, and 9. These differences themselves increase by 2, indicating a quadratic relationship. We can derive the nth term formula as ( a_n = n^2 + 8n + 1 ). Thus, the nth term of the sequence can be expressed as ( a_n = n^2 + 8n + 1 ).
To find the nth term of the sequence 9, 12, 17, 24, 33, 44, we first observe the differences between consecutive terms: 3, 5, 7, 9, 11. These differences form an arithmetic sequence with a common difference of 2. This suggests that the nth term can be expressed as a quadratic function. By deriving the formula, the nth term is given by ( a_n = n^2 + 8n - 1 ).
3n
12 - 5(n-1)
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) ] .