a (sub n) = 11 + (n - 1) x d
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.
If you mean -1 3 7 11 15 then the nth term is 4n-5 and so the next term will be 19
It is: nth term = 5-4n and so the next term will be -19
One of the infinitely many possible rules for the nth term of the sequence is t(n) = 4n - 1
This is an arithmetic sequence which starts at 14, a = 14, and with a common difference of -1, d = -1. We can use the nth term formula an = a + (n - 1)d to get an = 14 + (n - 1)(-1) = 14 - n + 1 = 15 - n.
a (sub n) = 11 + (n - 1) x d
3n - 1
20 - (3 * (n - 1))
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 of the sequence is 2n + 1.
If you mean -1 3 7 11 15 then the nth term is 4n-5 and so the next term will be 19
It is: nth term = 5-4n and so the next term will be -19
42
35 * * * * * That is the next term. The question, however, is about the nth term. And that is 6*n - 1
The nth term is 4n-1 and so the next term will be 19
Tn = 10 + n2