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Question : A $10$-year loan of $\$500$ is repaid with payments at the end of each year. The lender charges interest at an annual effective rate of $10\%$. Each of the first ten payments is $150\%$ of the amount of interest due. Each of the last ten payments is $X$. Calculate $X$.

My Attempt

$\$ 500$ will earn $\$50$ interest each year, so each of the first $10$ payments must be $\$75$.

Then after $10$ years, a total of $750$ has been repaid. In $10$ years, I can find the accumulated debt by saying

PV=500
I/Y=10
N=10

giving me FV=$\$1296.87$.
So the balance would be $\$ 1296.87-\$ 750=\$ 546.97$.

Now I am stuck! How do I find out what the last payments should be? I know I can't just divide $\$ 546.97/10=\$ 54.697$, because the lender is still charging interest while the borrower pays off this remaining debt, so there would still be the interest left over.

This situation isn't mentioned anywhere in my calculator manual! Can one of you give me some explanation about what is going on so that I can do it by hand?

I tried working on it some more, and came up with a really great idea! Since the payments are $1.5\times$ the $10\%$ interest, it's just like paying off $5\%$ of the principal each year! This saves me a lot of time, because I can just set $I/Y=-5$ and get $FV=598.74$ on my calculator. I did it the long way by calculating the future value of each interest payment (turns out they were not all $\$75$, because the outstanding principal got smaller), and they were the same. Is this always going to work, or did I just get lucky here?

Another update!
I think I solved it. All I needed to do was to set

 FV=0 
 I/Y=10 
 N=10 
 PV=598.74

and then I got $PMT=97.44$. I never used the PMT button before, though, so is there some other way I can check the answer is right?

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  • $\begingroup$ I like that this question explains what they've tried and where they got stuck. $\endgroup$
    – Noah Snyder
    Jul 24, 2010 at 2:00
  • $\begingroup$ I just deleted some of my previous comments as they might not have been correct. You said that each of the first 10 payments must be 150% of interest due and used that to get that each payment must be \$75 and the calculation in my answer depends on that figure. However, the amount of interest due decreases as payments are made, so each subsequent repayment will be less than \$75. The correct way to deal with this is to just reduce the debt by 5% per year $\endgroup$
    – Casebash
    Jul 24, 2010 at 3:46

1 Answer 1

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This problem is in two stages. For the first stage, notice that you are paying 150% interest, but ending up owing more. This is because you subtracted $\$ $750 from the future value, when in fact each $\$ $75 amount was paid at a time in the past and needs be converted to a future value too. The payment in the nth year has a future value of $75\times(1.1)^{10-n}$. The total future value of the repayment is: $$75(1.1^9)+75(1.1^8)+\cdots+75(1.1^0).$$

Note that I have assumed that the interest is charged before the repayments are made. This sequence is a geometric progression. We consider it as a geometric sequence in reverse to make the maths easier. It has first term ($a$) 75, each term 1.1 times the previous ($r$) and 10 terms ($n$). The sum is given by the formula: $$\frac{a{r^{n-1}}}{r-1}=\frac{75{1.1^{10-1}}}{0.1}\approx\$1195.31$$

After we have solved this first part, then it is just a standard interest with repayments problem..

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  • $\begingroup$ Order doesn't matter here, because interest is accrued continuously. Sorry I don't know how to find the rate, but its equivalent to 10% a year. The payments are once a year though. I do see what you mean about the future value of the payments, but its a pain to type all 10 into my calculator. Is there an easier way? I don't want to run out of time when I see a 30 yrs problem on the real exam! $\endgroup$
    – Larry Wang
    Jul 24, 2010 at 1:41
  • $\begingroup$ @Kaestur: Re order: We can make this problem discrete by assuming all the interest is charged an equivalent amount at the end of the year just before the payments. So, in this case, the repayments are made after the interest. Re: calculating the future value, this is a geometric series, not an arithmetic as I first said. There is a formula for summing this up, I'll update my question with more information $\endgroup$
    – Casebash
    Jul 24, 2010 at 1:55
  • $\begingroup$ @Casebash: The geometric series page is so confusing to me! But I think I found a simple way to do it, and added it to my question. I also calculated the balance after 10 years by discounting the first 10 payments, like you originally suggested. I didn't get the $1195 answer like you did, though. What happened? That would mean the balance after 10 years is only 101 dollars. That sounds too low if you are only paying back 1.5x the interest rate. $\endgroup$
    – Larry Wang
    Jul 24, 2010 at 3:11
  • $\begingroup$ @Kaestur: Wait, so you know complex analysis, but not geometric series? Are you seeding? $\endgroup$
    – Casebash
    Jul 24, 2010 at 4:01
  • $\begingroup$ @Casebash: I've introduced myself in my moderator nomination. This question was intended as an example of how a non-mathematician might ask a question on this site, and demonstrate things like showing what work you've done and interacting with people trying to solve your question. I'm sorry if you felt cheated by answering my fake question. I have upvoted you to thank you for the effort though :) $\endgroup$
    – Larry Wang
    Jul 24, 2010 at 4:25

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