What's the nth term for this sequence?

Notice that a is simply sequential so it will be "n". The b term increases by two each time and starts at 8 so 2n+6. Finally, c is sequential beginning at 7 so n+6.

[n][(2n+6) - (n+6)]

Well, if you look at the first four, you can clearly see that (A(B-C)) A is going up by one each time, C is going up by one each time, and B is going up by two each time. So I'd just write them all out until you reach the ninth one. (I'm assuming by nth you mean ninth, but I don't know for sure if that's what you mean...)

1(8-7)

2(10-8)

3(12-9)

4(14-10)

5(16-11)

6(18-12)

7(20-13)

8(22-14)

9(24-15)

1(1)=6, 2(2)=4, 3(3)=9, 4(4)=16, 5(16-11)=5(5)

it would be n(n) which is n^2

because the difference of b and c equates to a, the value inside and the value outside is equal. So for instance the 100th term is

100(b-c) you know b-c is 100, which would just be 100^2

5(16-11)

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n((6+2n)-(n+6)) = n²

n({6 + 2n} - {6+n} ), n=1,2,3,...{or n², n=1,2,3,...}