Loan Repayment

Relevant wiki: Time value of money

Let the interest rate be $i$ and the number of period or payments be $n$ . Then the future value $A_0$ after the $n$ periods is $\text{FV}_{A_0} = A_0(1+i)^n$ . We note that payment instalment $B$ has a periof $n-1$ , $n-2$ , $n-3$ , $\cdots$ , 2, 1, 0. Therefore, the future value of all $B$ 's is $\displaystyle \text{FV}_B = B\sum_{k=0}^{n-1}(1+i)^k$ . Since the payments with interest pay for the principal with interest,

$\begin{aligned} \text{FV}_B & = \text{FV}_A \\ B\sum_{k=0}^{n-1}(1+i)^k & = A_0(1+i)^n \\ \frac {B\left((1+i)^n-1\right)}i & = A_0(1+i)^n \\ \implies B & = \frac {A_0i(1+i)^n}{(1+i)^n-1} & \small \color{#3D99F6} \text{Putting the values} \\ & = \frac {500\cdot 0.2 \cdot 1.2^6}{1.2^6-1} \\ & \approx \boxed{150.35} \end{aligned}$

The dynamics of the outstanding balance are governed by the first-order difference equation:

$A_{k} = (1 + i) A_{k-1} - B$

where $i$ is the interest rate (as a fraction) and B is constant. The solution of the difference equation is

$A_{k} = c_1 (1+i)^k + c_2$

Using the method of undetermined coefficients, we simply plug in this solution into the difference equation to determine $c_1$ and $c_2$

$c_1 (1+i)^k + c_2 = (1 + i) ( c_1 (1+i)^{k-1} + c_2 ) - B$

the terms containing $c_1$ cancel out , and we get

$i c_2 = B$

So that,

$A_{k} = c_1 (1 + i)^k + \dfrac{B}{i}$

The initial condition is $A_0 = A$ the loan amount, hence,

$A = c_1 + \dfrac{B}{i}$

leading to,

$c_1 = A - \dfrac{B}{i}$

Thus we have the dynamics now fully specified, namely,

$A_{k} = (A - \dfrac{B}{i} ) (1 + i)^k + \dfrac{B}{i}$

Now we can apply the boundary condition, that $A_{n} = 0$ where $n$ is the number of months of the loan agreement.

Hence,

$A_{n} = 0 = (A - \dfrac{B}{i} ) (1 + i)^n + \dfrac{B}{i}$

Solving for $B$ , we obtain,

$\large B = A \dfrac{i (1 + i)^n}{ (1 + i)^n - 1 }$

Pluggin in the given values of $A = \$500$ and $i = 20\% = 0.2$ and $n = 6$ , we obtain,

$\large B = 500 \dfrac{ (0.2)(1.2)^6 }{ (1.2)^6 - 1 } = \$150.35$