What is the nth term of 2 3 7 14 24?

Answer 1

The nth term for n>5 can be any number you want as infinitely many polynomials can be found with the first 5 terms as {2, 3, 7, 14, 24} and (at least one of them) will give whatever values you choose for the nth terms with n>5.

However, the simplest sequence is given by the simplest polynomial which is a quadratic since the second difference is constant:

Seq: 2 3 7 14 24

1st dif: 1 4 7 10

2nd dif: 3 3 3

We are looking for a polynomial of the form t(n) = ax² + bx + c

a = ½ the second difference = ½ × 3 = 3/2

To avoid fractions whist working out a solution, we can double everything and then we want half the final quadratic:

Create the sequence of 3n² and subtract it from the original sequence (doubled) and find the common difference in the new sequence:

seq: 4 6 14 28 48

3n²: 3 12 27 48 75

sub: 1 -6 -13 -20 -27

diff: -7 -7 -7 -7

Thus the second coefficient is -7

Giving the quadratic so far as 3n² - 7n + c

Plugging in n = 1, gives: 3 - 7 + c = 4 → c = 8

Thus the doubled sequence is 3n² - 7n + 8

And the nth term of the original sequence is given by half this:

t(n) = ½(3n² - 7n + 8)

which can be expanded and written as:

t(n) = 3/2 n² - 7/2 n + 4

Answer 2

Any number that you choose can be the nth number. It is easy to find a rule based on a polynomial of order 5 such that the first five numbers are as listed in the question. There are also non-polynomial solutions. 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.
One possible solution, is the quartic polynomial, U(n) = (-11*n^3 +96*n^2 -205*n +132)/6 for n = 1, 2, 3, ...

Answer 3

28