1) What is the recurrence relation? i.e. if $x_{n}$ is owed at the end of the $n$th year, with $x_{0}=150000$, what is $x_{n+1}$ in terms of $x_{n}$? You should be able to work this out.
> It's $x_{n+1}=1.06x_{n}-10000$.
2) Can you use this to come up with a closed form expression for $x_{n}$?
Hint 1:
> Try $x_{n}=a\cdot1.06^{n}+c$. Substitute into the recurrence relation and then substitute in $x_{0}$ to get the two constants.
Answer:
> $x_{n+1} = 1.06x_{n}-10000 = 1.06\cdot a\cdot1.06^{n}+1.06c - 10000 = a\cdot1.06^{n+1}+1.06c-10000 = a\cdot1.06^{n+1}+c \Rightarrow 0.06c = 10000 = 10^{6}/6 \\\ x_{0}= 150000 = a + 10^{6}/6 \Rightarrow a = (900000-10^{6})/6 = -10^{5}/6 \\\ \therefore x_{n} = \frac{10^{5}}{6}\left(10 - 1.06^{n}\right)$
> You can find the number of years to pay the mortgage by finding the smallest integer $n$ for which $x_{n}\le 0$, which is $\lceil\ln{10}/\ln{1.06}\rceil = 40$.