If look for the pattern in these numbers, you can see that they follow a simple sequence in which each number is four more than the previous one. They are not multiples of four though, so they will have to be offset. In this case, that offset is two. This means that we can express this sequence as:
f(n) = 2 + 4n
assuming that 6 is the first term, with an "n" value of 1.
The sequence has a difference of 10, so the nth term starts with 10n. Then to get to -8 from 10 you need to subtract 18. So the nth term is 10n - 18.
The nth term in this arithmetic sequence is an=26+(n-1)(-8).
Well, isn't that just a lovely pattern we have here? Each term is increasing by 4, isn't that delightful? So, if we want to find the nth term, we can use the formula: nth term = first term + (n-1) * common difference. Just like painting a happy little tree, we can plug in the values and find the nth term with ease.
If the term number is n, then the nth term is 10(n-1) +8.
The nth term is 18 -3n and so the next term will be 3
1,7,13,19
The sequence has a difference of 10, so the nth term starts with 10n. Then to get to -8 from 10 you need to subtract 18. So the nth term is 10n - 18.
The nth term in this arithmetic sequence is an=26+(n-1)(-8).
t(n) = (-5n4 + 62n3 - 247n2 + 334n - 36)/6
The sequence given is -2, -8, -18, -32, -50. To find the nth term, we first observe the differences between consecutive terms: -6, -10, -14, -18, which show that the second differences are constant at -4. This indicates that the nth term can be expressed as a quadratic function. By fitting the sequence to the form ( a_n = An^2 + Bn + C ), we find that the nth term is ( a_n = -2n^2 + 2n - 2 ).
It is: 25-7n
To find the nth term of the sequence -2, -8, -18, -32, -50, we first observe the differences between consecutive terms: -6, -10, -14, -18. The second differences (which are constant at -4) suggest that the nth term can be represented by a quadratic function. The general form is ( a_n = An^2 + Bn + C ). Solving for coefficients A, B, and C using the first few terms gives the nth term as ( a_n = -2n^2 + n ).
Un = 25 - 7n
Well, isn't that just a lovely pattern we have here? Each term is increasing by 4, isn't that delightful? So, if we want to find the nth term, we can use the formula: nth term = first term + (n-1) * common difference. Just like painting a happy little tree, we can plug in the values and find the nth term with ease.
To find the nth term of the sequence 4, 10, 18, 28, 40, we first identify the pattern in the differences between consecutive terms: 6, 8, 10, and 12. The second differences are constant at 2, indicating a quadratic sequence. The nth term can be expressed as ( a_n = n^2 + n + 2 ). Thus, the nth term of the sequence is ( n^2 + n + 2 ).
If you mean: 6 12 18 24 then the nth term is 6n
To find the nth term of the sequence 5, 15, 29, 47, 69, we first determine the differences between consecutive terms: 10, 14, 18, and 22. The second differences are constant at 4, indicating that the nth term is a quadratic function. By fitting the quadratic formula ( an^2 + bn + c ) to the sequence, we find that the nth term is ( 2n^2 + 3n ). Thus, the nth term of the sequence is ( 2n^2 + 3n ).