# Financial mathematics

I was given the following information:

A young couple decide that they want to buy a house for 220000. They have saved a deposit of 80000 and are confident that they can pay 2500 per month, starting in 3 months' time to amortise their loan. The current interest rate is 20% p.a., compounded monthly.

The question:

What will the amount of the last payment (less than 2500) be?

Now I thought all payments would be 2500, if the loan is to be amortised?

• They are confident that they can pay that...that doesn't mean that is what their payment will be...also, what is the term of the loan? Amoritization determines the amount of interest and principle in each payment. Oct 13, 2013 at 14:43
• Also, If they actually pay that much, there will be a residual payment as the final payment depending on the interest rate...that is what they are looking for. Oct 13, 2013 at 14:45
• My book defines amortisation in the following way: When a loan is repaid in equal installments, we say that the loan is amortised over a time period. Oct 13, 2013 at 14:57
• How would I attempt to find the last payment? Oct 13, 2013 at 14:57
• Have you heard of a "drop payment"? What is the term of the loan? Just calculate the present value with equal installments of 2500, and a final payment of X. Solve for X. Oct 13, 2013 at 15:03

If $P$ is your payment and $L$ is the loan amount, then $$L=P\left(\frac{1-\frac{1}{(1+i)^n}}{i}\right)$$

You are solving for $n$, since you already know $L, P,$ and $i$. $n$ will tell you the term of the loan, but it will most likely not be a natural number.

Then set

$$L_n=P\left(\frac{1-\frac{1}{(1+i)^{\lfloor{n}\rfloor}}}{i}\right)$$ (I substituted $\lfloor{n}\rfloor$ back in the expression).

Your drop payment should be $L-L_n$

Note, you have to convert your annual percentage to monthly. Use the following conversion: $$(1+i)^n=\left(1+\frac{i^{(12)}}{12}\right)^{12n}$$

Finally, it says that you are waiting 3 months to begin payments so assuming interest is accruing, your initial $L$ should be $140,000(1+.0153095)^3=146,528.932.$

(since your monthly interest using the above formula is $1.53095\%$)

So you have this expression $$146,528.095=2,500\left(\frac{1-(1.0153095)^{-n}}{.0153095}\right)$$ Solving for $n$ you get $n=149.806$. This tells us that our $150$th payment is not a full payment. Thus, $\lfloor{n}\rfloor=149$. Plugging this into the right hand side and solving for $L_n$ give us $$L_n=2,500\left(\frac{1-\frac{1}{(1.0153095)^{149}}}{.0153095}\right)=146,322.291$$ Finally, $L-L_n=146,528.932-146,322.291=206.641$

Hence, your final payment is $\$206.641\$.