I assume that sequence is 3,4,6,9,13,18,24
delta y=1,2,3,4,5,6
delta delta y=1,1,1,1
So this sequence has a constant acceleration, which is the property of a parabola or quadratic equation of the form ax^2+bx+c=y. Using three data points to create a system of equations we can solve for the three variables.
9a+3b+c=6, 4a+2b+c=4, a+b+c=3, getting differences
5a+b=2, 3a+b=1, and getting differences again
2a=1, a=1/2, making 5a+b=2 become:
2.5+b=2, b=-1/2, making a+b+c=3 become:
1/2-1/2+c=3, c=3
So the quadratic equation that produces that sequence is:
y=x^2/2-x/2+3 or more neatly
y=(x^2-x+6)/2, so the 8th term is:
y(8)=(8^2-8+6)/2
y(8)=(64-8+6)/2
y(8)=62/2
y(8)=31
