It could be any number you like.
You can select a polynomial of order 5 so that the nth term is any number at all.
The simplest solution for the above sequence, though is Un = 5n - 16
11
The nth term is: 3n-7 and so the next number will be 11
The nth term is: 3n-7 and so the next number will be 11
3n-2
1...2....3....43...7...11..157 - 3 = 411 - 7 = 415 - 11 = 4(the start of the nth term is 4n)4 x 1 = 4(but the first term is 3, so...)4 -1 = 3nth term = 4n - 1
To find the nth term of the sequence -4, -1, 4, 11, 20, 31, we first identify the pattern in the differences between the terms: 3, 5, 7, 9, 11, which increases by 2 each time. This suggests a quadratic relationship. The nth term can be expressed as ( a_n = n^2 + n - 4 ). Thus, the nth term of the sequence is ( a_n = n^2 + n - 4 ).
The 'n'th term is [ 4 - 3n ].
The 'n'th term is [ 4 - 3n ].
The 'n'th term is [ 4 - 3n ].
It works out as -5 for each consecutive term
It is 5n-4 and so the next term will be 21
1 4 9 is a series of squared numbers. The nth term is [ n squared ]