Help with finding the nth term!

Hey I'm in algebra 2 this year, and I'm supposed to be able to determine what type of sequence the following are, and then complete the problem. If anyone could explain how to do any of these to me, that'd be really helpful!

1. a = -5, d=4, n=9; find the nth term

2. a=5, n=4, r=3; find the nth term

3. a=3, d = -4, n=6; find the nth term

4. a = -4, n=6, r = -2; find the nth term

a stands for "a sub 1" or the 1st term of the sequence. d stands for the common differnce (only in arithmetic sequences). r stands for the common factor (only in geometric sequences). So, the easiest way to do this is to write the expression for the nth term and then plug in the value for n.

For a geometric sequence, the standard form of the nth term expression is (I couldn't find a sub-n, so pretend the sub-x is an n. Also couldn't find a subscript 1, so pretend the ᵦ is 1):
aₓ=aᵦ(r)^(n-1)

For an arithmetic sequence, the standard form of the nth term expression is (I couldn't find a sub-n, so pretend the sub-x is an n. Also couldn't find a subscript 1, so pretend the ᵦ is 1):
aₓ=aᵦ+d(n-1)

So that's all you need to know. (The expressions make it look really complex, but really it's almost just common mathematical sense)

1. There is a d and no r, so it must be arithmetic.
aₓ=aᵦ+d(n-1)
aₓ=-5+4(n-1)
Simplify that to:
aₓ=4n-9
And that's the nth term expression. Now, in this case, you are looking for the 9th term, so:
aₓ=4(9)-9
=25
Follow the same steps for the rest.
2. aₓ=5(3)^(n-1)
aₓ=5(3)^3
aₓ=5(27)
=135
3. aₓ=-4n+7
aₓ=-4(6)+7
=-17
4. aₓ=-4(-2)^(n-1)
aₓ=-4(-2)^5
aₓ=-4(-32)
=128