$0.1$% 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 $n$.
$A(2)$ is the amount at the end of the second month. A deposit of $x$ may made at the beginning of the first month, so after two months it has grown to $1.004^2x$. A deposit of $x$ was also made at the beginning of the second month. At the end of the second month it has grown to $1.004x.$ So $$A(2)=1.004^2x+1.004x$$
Does it make sense now?
Best Answer
The total value of all the payments and the accumulated interest, at $i=0.0025,$ is given by$$100+100(1+i)+100(1+i)^2+100(1+i)^3+...+100(1+i)^{n-1}$$The last payment accumulates zero interest, and the first payment compounds for $n-1$ months. The sum of this geometric series, $S$, is given by:$$S=\frac{100((1+i)^n-1).}{i}$$ You then need to subtract $100n$ from this total to find the interest portion.