SOLUTION: Nth term of -7,-6,-3,2,9,18,29

List in a column. Subtract each number
from the number under it, and place this
difference to the right of each number in
a new column to the right:

-7  1
-6  3
-3  5
 2  7
 9  9
18 11
29

Do the same with the second column column

-7  1  2
-6  3  2
-3  5  2
 2  7  2
 9  9  2
18 11
29

It takes 2 additional columns to get a column that contains only
one number.  So we expect a 2nd degree equation for the
nth term.  So we will assume an equation of this form:

nth term = tn = An² + Bn + C 

Then we substitute 1,2 and 3.

1st term = t1 = A(1)² + B(1) + C = A + B + C = -7

2nd term = t2 = A(2)² + B(2) + C = 4A + 2B + C = -6

3rd term = t3 = A(3)² + B(3) + C = 9A + 3B + C = -3
 
So we have this system of 3 equations in 3 unknowns:

 A +  B + C = -7
4A + 2B + C = -6
9A + 3B + C = -3

Solve that system of equations and get A = 1, B=-2 and c=-6

So the nth term tn = An² + Bn + C becomes:

nth term = tn = n² - 2n - 6  

Edwin