We have \(f_n = 3f_{n-1}-60\) and \(f(1)=f_1=20\)

For a difference equation in the form \(f_n = af_{n-1}+c\) , the general solution takes form of: \(f_n=x+\beta a^n\) where x is the fixed point of the difference equation:

1) \(f(0)=f_0 = \frac{f_1+60}{3}=\frac{80}{3}\)

2) Finding the fixed point \(x\) -> Let \(f_n=f_{n-1}=f\), then:

\(f=3f+60 \quad=>\quad f=30\) so the fixed point \(x\) is 30.

3) To find \(\beta\) in the general solution, work out the case n=0:

\(f_0=\frac{80}{3}=30 +\beta 3^0 = 30+\beta*1 \quad=>\quad \beta=\frac{80}{3}-30=\frac{-10}{3}\)

4) Now we have the general solution just plug in whatever term you want to find, in this case the 10th term:

\(f_{n}=30-\frac{10}{3}*3^{n}\\f_{10}=30-\frac{10}{3}*3^{10} = -196800\)