Year 12 Finance - Combining Superannuation and withdrawals

1)
0.03 per annum / 12 = 0.0025 per month 
We have $$ M_n = M_{n-1}\times 1.0025 - 6000 = M_{n-1}r - 6000$$In other words, what remains after $n$ months is what we had after $n-1$ month with its monthly interest, minus the 6000 withdrawn. Since $M_0 = A$ we get $$M_1 =M_0r-6000=Ar-6000 $$ $$M_2 =M_1r-6000=(Ar-6000)r-6000= Ar^2-6000(1+r)$$ $$ M_3 =M_2r-6000=(Ar^2-6000(1+r))r-6000= Ar^3-6000(1+r+r^2)$$ and so we inductively get $$M_n =Ar^n-6000(1+r+\dots +r^{n-1})=Ar^n-6000{r^n-1\over{r-1}} $$ Now note that $$ 2400000=6000/0.0025=6000/(r-1)$$ So $$ M_n =Ar^n-2400000(r^n-1)=2400000-(2400000-A)r^n$$ as desired.

2) 
If $M_n=0$ then she cannot withdraw anymore. So we get $$2400000-(2400000-A)r^n=0 \\ \implies 2400000=(2400000-A)r^n =647418.82\times r^n$$ So $$ r^n =3.707$$ and $$ n\log r =\log 3.707$$ giving $$n=524.747 $$But n must be an integer, so after 524 months there is less than 6000 in the account and no further withdrawals is possible. In fact $$M_{524} =4472.7$$