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?
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 $\$206.641$.