There are infinitely many polynomials of order 6 (or higher) that will give these as the first six numbers and any one of these could be "the" rule. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.
In this particular case, the simplest solution isU(n) = 3*n^2 - 1 for n = 1, 2, 3, ...
134
jj
10
It is: nth term = 7n-9
foot
(13 + 15 + 26 + 11 + 26 + 16 + 12)/7 = 17 ====
2 5 10 17 26 37 50 65 82 (3+5+7+9+11...)
First put them in order:11, 12, 13, 15, 16, 26, 26.Then use your fingers the numbers at the opposite ends:11, 12, 13, 15, 16, 26, 26.Then keep moving in11, 12, 13, 15, 16, 26, 26.11, 12, 13, 15, 16, 26, 26.Until you come to the middle:11, 12, 13, 15, 16, 26, 26.So the median is 15.
The nth term in this arithmetic sequence is an=26+(n-1)(-8).
It is: nth term = 7n-9
It is: nth term = 6n-4
It is: nth term = 35-9n
Tn = 10 + n2
46n9
It is: 26-8n
Here are the first five terms of a sequence. 12 19 26 33 40 Find an expression for the nth term of this sequence.
[ 6n + 8 ] is.
It is: 26-6n
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......
t(n) = 4n2 - 4n + 2