No number completes it because the sequence can go on forever. There are infinitely many possible solutions: given any seventh number it is possible to find a polynomial of degree 6 that will fit the above six points and the selected seventh.
There is only one polynomial of degree 5, and using that as the position to value rule, the next number is 131.
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The number which best completes the sequence below is:2 9 5 13 10 19 17?
The number that completes the sequence would be 11. Imagine the sequence as two sequences of every other number starting with the first 2 and the first 6. You would have 2 3 6 and 6 8 12 2 3 6 is adding odd numbers starting with one. 6 8 12 is adding even numbers starting with two. The next number belongs to 2 3 6 so you would add 5 to get 11.
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