A bank offers loans at interest of $12\%$ per annum , compounded monthly. Jack took a $\$50,000$ loan and makes a monthly installment payments of $\$4200$ at the end of each month . Find his outstanding loan after his third installment payment.
I'm confused with what to substitute into the compound interest formula:
$$ A = 50,000 \left( 1 + \frac{0.12}{12} \right)^n $$
Since I do not know what's the total number of interest periods, which number should I substitute into $n$ ?
On
Let a month be your period, so the answer to your actual question is that you should substitute 3 into n, because 3 is the number of periods. However, the interest rate is not $0.12\%/12$
To find the monthly interest rate you must first use $n=12$ and put $12\%$ on the left hand side of your equation in order to calculate the monthly interest rate $x$:
$112\%=(1+x)^{12}$
$x=\sqrt[12]{1.12}-1 = 0.9489\%$ per month$
Now just use brute force to calculate what he owes each month by applying this algorithm 3 times to the principal $\$50,000$: Add $x\%$ interest and deduct $\$4,200$.
After 1 month he owes $\$46,274.44$
After 2 months he owes $\$42,513.53$
After 3 months he owes $\$38,716.93$
On
While the solutions to approach this through "brute force" are fine for a small number of periods, I thought I'd show a more general approach when you're asked to do the computation for longer durations. If you find the following heavy-going, then just do the brute force computation (I indicate where you can immediately do this in the working below).
Approach this problem from first principles. An annual interest rate of $12\%$ compounded monthly is a compound monthly rate of $1\%$. A monthly compounding means the amount owed will grow by a factor of $(1 + 0.01) = 1.01$ with each month.
Let the initial amount at the beginning of the first month (the principal) be represented by $A_0$.
Then the amount at the end of the first month $A_1 = 1.01A_0 - 4200$.
The last term is because he does a partial repayment by $\$4200$.
In fact, this is the general formula for the recursion:
$A_{n+1} = 1.01A_n - 4200$.
At this point, you can immediately apply this recursive calculation to find $A_3$, which is the simple "brute force" approach. Use $A_0$ to find $A_1$, then $A_1$ to find $A_2$, then finally $A_2$ to find $A_3$.
But if you want a more general solution, you can work through this to find a closed form for $A_n$.
$A_n = 1.01A_{n-1} - 4200 = 1.01^2A_{n-2} - 4200(1.01 + 1) \\= 1.01^3A_{n-3} - 4200(1.01^2 + 1.01 + 1) \\= \dots \\= 1.01^nA_0 - 4200(1.01^{n-1} + 1.01^{n-2} + \dots + 1.01^0)$
the last term of which can be further simplified by the geometric sum formula.
$$4200(1.01^{n-1} + 1.01^{n-2} + \dots + 1.01^0) = 4200\frac{1.01^n-1}{1.01-1} = 4.2 \times 10^5(1.01^n-1)$$
After a little rearrangement, you will get:
$A_n = 4.2 \times 10^5 - 1.01^n(4.2 \times 10^5 - A_0)$
This closed form can be proved using induction for rigour.
From this, you can find out $A_3 = 4.2 \times 10^5 - 1.01^3(4.2 \times 10^5 - 50000) = \$38,788.63$. This is the answer to your question.
If you pursue this to the end (by setting $A_n = 0$), you will find he will have fully repaid his loan by the $13$th month.
Do this step-by-step. Jack takes out a $\$50,000$ loan. One month goes by, how much does he owe? Use the compound interest formula: $$ 50000\left(1+\frac{0.12}{12}\right)^{12t} $$ where $t$ is in years. So after 1 month, how much does he owe$?$ To do this, plug in for $t$ being $1$ month taking careful note of that $t$ is in years! Then subtract off the amount Jack pays that month. This new number is the amount he owes that will accrue more interest in the next month. Do this same procedure for the next month and then the next month, each time using the new amount. After that, you will know exactly how much Jack owes! Aren't you glad they didn't ask you how much he owes after $12$ months?!$