In your first equation, the difference of the subscripts should be 1, not 2. If $A*$ is the limit, you are seeking a value of the account that doesn't change. From $A*=1.03A*-15,$ you add $15$ to each side and subtract $A*$ from each side to get $15=0.03A*$, then divide by $0.03$ to find $A*$. But this doesn't get the value after 10 years, it gets the value that would be stable. In fact, the account is growing. The first year you get 30 in interest and pay 15 in fees, so $A_1=1015$ and it will grow by more than \$15 each year thereafter.
Added: as this is a linear recurrence, the full solution is a particular solution to the full equation plus any solution to the homogeneous part. You have found $A*=500$ regardless of $n$ is a solution to the whole recurrence without the initial condition. Now the homogeneous part $A_{n+1}=1.03A_n$ has the solution $A_n=a1.03^n$ for any constant $a$, so the general solution is $A_n=500+a1.03^n$ Given that $A_0=1000$, we can solve to find $a=500$ and the full solution is $A_n=500+500\cdot1.03^n$
Added 2: You could check out Wikipedia, though it may be too complex. Your basic equation looks like $A_{n+1}=rA_n+C$, which is called a linear recurrence because none of the $A$'s are multiplied by each other-they can be multiplied by constants or functions of $n$ and the equation is still linear. It is convenient to also consider the homogeneous part of the equation, with all terms that do not involve $A$'s, which here is $A_{n+1}=rA_n.$ The point is that if you add any solution to the homogeneous equation to a solution of the original equation, it is again a solution. Let us suppose $f(n)$ satisfies the original equation, so $f(n+1)=rf(n)+c$, and $g(n)$ satisfies the homogeneous part, so $g(n+1)=rg(n)$. Also any constant times $g(n)$ satisfies the homogeneous part. Then $f(n+1)+kg(n+1)=rf(n)+c+rkg(n)=r(f(n)+kg(n))+c$. So to find all solutions to the original equation, we can find any solution (called a particular solution) to the original equation and add to it all solutions of the homogeneous equation. An easy particular solution to find is to assume it does not depend on $n$, which is your $A*$. If we just substitute $f$ (instead of $f(n)$) into the original equation, we find $f=\frac{c}{1-r}$, which is where your $A*=\frac{-15}{-.03}=500$ came from-it is the constant solution to your equation. To find all solutions to the homogeneous equation, we have $A_{n+1}=rA_n$. The solutions can be found be inspection and are $dr^n$ for some constant $d$. So the general solution is $A_{n}=dr^n+\frac{c}{1-r}=d1.03^n+500$. To find $d$, you use your initial condition. Here we are given that $A_0=1000=d1.03^n+500=d+500,$ so $d=500$ and the full solution is $A_n=500+500\cdot1.03^n$
Your first deposit compounds for 120 months.
The second for 119 months.
Your last deposit for 1 month.
The future value of 120 deposits.
$10(1.005)^{120} + 10(1.005)^{119} + \cdots + 10(1.005)$
or
$\sum_{n=1}^{120} 10(1.005)^n$
This is the sum of a geometric progression.
$\sum_{n=1}^{m} y^n = \frac {y(y^m-1)}{y-1}$
to find this formula multiply by $\frac {1-y}{1-y}$
$\frac {1}{1-y}(1-y)(y+y^2 + y^3 + \cdots + y^m) = \frac {1}{1-y} (y - y^2 + y^2 -y^3 + y^3-y^4 + \cdots - y^{m+1})$
The expression "telescopes" leaving:
$\frac {1}{1-y} (y - y^{m+1})$ which equals the formula above.
plugging the numbers for this problem.
$10\frac {1.005(1.005^{120} - 1)}{0.005}$
Best Answer
Here is an answer done "their" way- We have $$1300=500(1.06)^n$$ $$\implies 13=5(1.06)^n$$ $$\implies \frac{13}{5}=(1.06)^n$$ $$\implies \log_{1.06}\frac{13}5 =n$$ $$\implies n \approx 16.39830702$$ $$\implies n \approx 16.4$$ Rounded to one decimal place.
Please ask if you need any further clarification.