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.
t(n) = 4n2 - 4n + 2
46n9
Tn = 10 + n2
[ 6n + 8 ] is.
It is: nth term = 7n-9
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.
It is: nth term = 35-9n
t(n) = 4n2 - 4n + 2
The common difference (d) between successive terms is -9. The first term (a) is 26 The formula for the nth term [a(n)] of an Arithmetic Series is , a + (n - 1)d. Inputting the values for a and d gives :- a(n) = 26 - 9(n - 1) = 26 - 9n + 9 = 35 - 9n......where n = 1,2,3......
46n9
Tn = 10 + n2
[ 6n + 8 ] is.
It is: 26-6n
t(n) = 10 - 6n where n = 1, 2, 3, ...
To find the nth term of the sequence 11, 21, 35, 53, 75, 101, we can observe the differences between consecutive terms: 10, 14, 18, 22, and 26, which increase by 4 each time. This suggests that the sequence can be described by a quadratic function. The nth term can be represented as ( a_n = 5n^2 + 6n ), where n starts from 1. Thus, the nth term corresponds to this formula for values of n.
The simplest rule is Un = 26 - 8n for n = 1, 2, 3, ... but it is always possible to fit a polynomial of degree 5 to the given sequence of numbers along with any sixth number.