Sequence Question....... help with this...........

Find the nth term of 12, 11, 9, 6.........

plz tell everything you do ........ be brief as possible

Note that 12 - 11 = 1, 11 - 9 = 2, 9 - 6 = 3. In other words, the successive differences are "1, 2, 3, ...", which appears to be an arithmetic sequence. This means that the original sequence is a quadratic sequence. (If you didn't already know this, it's a good fact to know.) In other words, the sequence has the form s(n) = a n^2 + b n + c for some constants a, b, c. Now we plug in the known values:

12 = s(1) = a + b + c,
11 = s(2) = 4a + 2b + c,
9 = s(3) = 9a + 3b + c.

It remains to solve this system of equations. Subtracting the second equation from the first, we get -1 = 3a + b. Multiplying this equation by 3, we get -3 = 9a + 3b. Subtracting this from the original third equation, we get 12=c. Plugging in 12=c, we now need to solve

0 = a+b,
-1 = 4a + 2b = 2(a+b) + 2a.

Since a+b=0, the second equation now implies a=-1/2. Finally, b = -a = 1/2. This proves that the nth term of the sequence is

s(n) = an^2 + bn + c = 12 - (1/2)(n^2 - n).

12
12 - 1
11
11 - 2
9
9 - 3
6
6 - 4
2
2 - 5
-3
-3 - 6
-9
-9 - 7
-16
-16 - 8
-24
-24 is now the 9th term
or
12
12 - 1
11
11 - 2
9
9 - 3
6
6 - 5
1
1 - 8
-7
-7 - 13
-20
-20 - 21
-41
-41 - 34
-75 could also be the 9th term if you were subtracting the fibonacci sequence from it.
idk either works, but i would go w/ the top one
Sorry if I wasn't any help

12, 11, 9, 6, 2, - 3, - 9, - 16, - 24.