There are two ways you can try to solve this problem. The first is try try to keep a running tab of how much is still owed at time $n$. For the sake of legibility, I am going to divide the amounts by 1000. The recursive formula you get is then:
$$a_0=150,\qquad a_{n+1}=(1+0.1)a_n-20.$$
If you compute the first few terms of this sequence, you get
$$a_1=150(1.1)-20, a_2=150(1.1)^2-20(1.1)-20, a_3=150(1.1)^3-20(1.1)^2-20(1.1)-20\ldots$$
We can factor and get a general formula
$$a_n=150(1.1)^n-20(1+1.1+1.1^2+\ldots+1.1^{n-1})$$
We have that $a_n$ is the difference of two terms: one is a geometric series, and the other is growing exponentially. We can use the formula for the sum of a geometric series to get
$$a_n=150(1.1)^n-20\frac{(1.1)^n-1}{0.1}.$$
However, we can get to this formula more quickly by reinterpreting what interest means. If you have investment opportunities, a dollar today is worth more than a dollar tomorrow, because you can invest that dollar to earn interest. If an investment of $1$ dollar today yields $1+i$ dollars tomorrow, then $1$ dollar tomorrow is worth only $1/(1+i)$ dollars today. We can then treat interest calculations as being akin to currency conversion calculations. The same way you might convert a dollar into some number of euros, when interest is computed you are converting today dollars into tomorrow dollars (or next year dollars, or whatever your interest periods are).
Now, we want to reinterpret the problem by converting all future dollar amounts into today-dollars. This has the advantage that the $150$ in our formulas never changes. It is just our initial amount measured in today dollars. However, the amount paid in future dollars has to be converted. Let $b_n$ be the amount still owed, but this time measured in today-dollars. The amount we pay back in year $n$ is then $20(1.1)^{-n}$, and so
$$b_n=150-20(1.1)^{-1}-20(1.1)^{-2}-\ldots -20(1.1)^{-n}=150-20(1.1)^{-1}\left(\frac{1-(1.1)^{-n}}{1-(1.1)^{-1}} \right).$$
If you clean up the expression slightly, you can verify that $a_n=b_n(1.1)^n$, which is what we would hope: to calculate the amount we owe at time $n$, we can do all our calculations in terms of today-dollars, and then convert today-dollars into time-$n$ dollars.
Regardless, we want to solve for the $n$ that makes $b_n=0$. We get
$$1-(1.1)^{-n}=3/4 \Longrightarrow (1.1)^{-n}=1/4 \Longrightarrow n=\log 4/\log 1.1\approx 14.55$$
However, this isn't a whole number, so we have to be careful in interpreting this "answer." Since we only pay back the loan once a year, the result tells us that we are not done after the $14$th year, but we are done by the $15$th year. Our final payment in the $15$th year will be less than the $20$ we pay every other year. To find out how much we need to reduce our final payment by, let us calculate $a_{15}\approx -8.86$. Thus, if we did pay 20 in the final year, we would have overpaid by $8.86$, and thus our final payment should be for $20-8.86=11.14$.
Best Answer
$0.1$% is $0.001$, so the monthly multiplier is $1.001$, not $1.1$.
If $n$ is the number of months, $a$ is the monthly deposit ($\$400$ in your problem), and $r$ is the multiplier ($1.001$ in your problem), the first deposit grows to $ar^{n-1}$, since it’s compounded $n-1$ times, the second to $ar^{n-2}$, and so on, down to the $n$-th (last) deposit, which ‘grows’ to $a$, since it’s not compounded at all. You therefore have $$a+ar+ar^2+\ldots+ar^{n-2}+ar^{n-1}\;,$$ which is indeed a geometric series. You almost remembered the formula for its sum:
$$a+ar+ar^2+\ldots+ar^{n-2}+ar^{n-1}=\frac{ar^n-a}{r-1}=\frac{a(r^n-1)}{r-1}\;.$$
In your problem that comes to $$\frac{400(1.001^n-1)}{1.001-1}=\frac{400(1.001^n-1)}{0.001}=400,000(1.001^n-1)\;.$$
If $C$ is the cost of the car, you want to solve $400,000(1.001^n-1)=C$ for $n$. After some basic algebra you have $$1.001^n=\frac{C}{400,000}+1\;;$$ if you now take logarithms on both sides you have $$n\log 1.001=\log\left(\frac{C}{400,000}+1\right)\;,$$ which is easily solved for $n$.