##### Assuming the first term of each sequence is n=1:

- \( a_n = 4-4n \) and \(a_{20} = -76\)
- \( a_n = \frac23(n-1)\) and \(a_{20} = \frac{38}{3}\)
- \(a_n = 36-11n\) and \(a_{20} = -184\)
- \( a_n= -5 + \frac72 n\)
- \( a_n=6 – \frac12 n\)
- \( a_n = -70+9n\)
- \( a_n = -5+9n\)
- \( a_n = 22-4n\)
- \(a_n = \frac{12}{5} – \frac25 n\)
- \( 2585\)
- \( n = 5\)
- \(a_n = 2\cdot 3^n\)
- \(a_n= \frac{15}{2} \cdot \left( \frac25 \right)^n \)
- \(a_n = 8 \cdot \left(\frac12\right)^n \)
- \(a_n = 60 \cdot \left(\frac12\right)^n \)
- \(a_n = -3072 \cdot \left( -\frac14\right)^n \)
- \(a_n = \frac45 \cdot 5^n \)
- \( a_n = \frac32 \cdot 2^n \) or \( a_n = \frac32 \cdot (-2)^n \)
- \( a_n = -160 \cdot 2^n\) or \(a_n = -160 \cdot (-2)^n\)
- \( a_n = 14\cdot \left(\frac12\right)^n\) or \(a_n = -14 \cdot \left(-\frac12\right)^n\)
- \( 5115 \) (Note, problem should read “…of the geometric series.”)

There’s also a chance that the teacher uses \( (n-1) \) more than \(n\) in the answers to 1-9 and 12-20. If so, then the teacher’s answers can be simplified into these answers.