The simplest, out of infinitely many possible answers, is the linear polynomial,
U(n) = 4n - 1 for n = 1, 2, 3, ...
The nth term = 4n-1
Un = 29 - 9n
The given sequence is 11, 31, 51, 72 The nth term of this sequence can be expressed as an = 11 + (n - 1) × 20 Therefore, the nth term is 11 + (n - 1) × 20, where n is the position of the term in the sequence.
One of the infinitely many possible rules for the nth term of the sequence is t(n) = 4n - 1
If by using the formula: nth term = n2+2 the sequence is 3 6 11 18 27 38 51 66.Then it follows that by using the same formula the sequence is III VI XI XVIII XXVII XXXVIII LI LXVI in Roman numerals.
The 'n'th term is [ 4 - 3n ].
Un = 29 - 9n
The given sequence is 11, 31, 51, 72 The nth term of this sequence can be expressed as an = 11 + (n - 1) × 20 Therefore, the nth term is 11 + (n - 1) × 20, where n is the position of the term in the sequence.
The nth term in this sequence is 4n + 3.
The nth term of the sequence is 2n + 1.
One of the infinitely many possible rules for the nth term of the sequence is t(n) = 4n - 1
I believe the answer is: 11 + 6(n-1) Since the sequence increases by 6 each term we can find the value of the nth term by multiplying n-1 times 6. Then we add 11 since it is the starting point of the sequence. The formula for an arithmetic sequence: a_{n}=a_{1}+(n-1)d
The nth term is 4n-1 and so the next term will be 19
3 11
The nth term of the sequence is (n + 1)2 + 2.
If by using the formula: nth term = n2+2 the sequence is 3 6 11 18 27 38 51 66.Then it follows that by using the same formula the sequence is III VI XI XVIII XXVII XXXVIII LI LXVI in Roman numerals.
It works out as -5 for each consecutive term
The 'n'th term is [ 4 - 3n ].