Section 1.3, Problems 2(e) and modified 3(a)

avalues= RecurrenceTable[{a[n+1]==-1.2a[n]+50,a[0]==30},
a,{n,0,20}]

modeldata=ListPlot[avalues]

In the above graph we are off by one unit in the horizontal axis. Rather than a=30 for n=0, or graph has a=30 for n=1.  A solution to this is to start with n=1 in the problem.

avalues= RecurrenceTable[{a[n+1]==-1.2a[n]+50,a[1]==30},
a,{n,1,21}]
modeldata=ListPlot[avalues]

Here is a solution that works well for simple equations.

a[0]=30;
a[n_]:=-1.2a[n-1]+50;
ListPlot[Table[{n,a[n]},{n,0,20}]]
Clear[a]

To keep starting n value zero. We can build our own ordered pairs. This works well for all types of equations.

nvalues=Table[n,{n,0,20}];
avalues= RecurrenceTable[{a[n+1]==-1.2a[n]+50,a[0]==30},
a,{n,0,20}];
points=Transpose[{nvalues,avalues}]
ListPlot[points]

nvalues=Table[n,{n,0,20}];
avalues= RecurrenceTable[{a[n+1]==-1.2a[n]+50,a[0]==22},
a,{n,0,20}];
points=Transpose[{nvalues,avalues}]
ListPlot[points]

Section 1.3, Problem 5 
Class Work

avalues= RecurrenceTable[{a[n+1]==1.005a[n]+200,a[0]==500},
a,{n,0,60}];nvalues=Table[n,{n,0,60}];
nvalues=Table[n,{n,0,60}];
points=Transpose[{nvalues,avalues}]

Log[4/3]/Log[1.005]