Edit: A farmer borrows $80,000 to purchase new machinery. The interest is calculated monthly at the rate of 2% per month, and is compounded each month.

The farmer intends to pay the loan with interest in two equal installments of $M at the end of the first and second years.

i) How much does the farmer owe at the end of the first month?

ii) Write an expression involving M for the total amount owed by the farmer after 12 months, just after the first installment of $M has been paid.

iii)Find an expression for the amount owed at the end of the second year and deduce that: $M = \frac{80000(1.02)^{24}}{(1.02)^{12} + 1}$

iv) What is the total interest over the two year period?

Any help will be appreciated, but I was particularly stuck with (ii) and (iii)


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The original loan was $\$80000$, and you probably figured out that after a month the farmers owes $\$80000\times 1.02$.

After 12 month, the total owed is $ \$80000\times 1.02^{12}$. At this point the farmers pays $M$, so the answer to (ii) is $ \$80000\times 1.02^{12}-M.$

At the start of the second year, the amount owed is the answer to part (ii). With interest, at the end of the second year, the amount is $ (\$80000\times 1.02^{12}-M)\times 1.02^{12}$. Remember that you were told that this is the same as $M$, since you have two equal payments. Therefore $$(80000\times 1.02^{12}-M)\times 1.02^{12}=M$$

Expanding the parenthesis, and moving the terms with $M$ on one side will yield the answer to part (iii)


Recall that the formula for compound interest is $A = P(1+i)^n$,


  • $A$ is the final amount
  • $P$ is the initial amount
  • $i$ is the interest rate (in decimals)
  • $n$ is the time that has elapsed.

For question (I):

Since the initial amount is $80 000$,

$A = 80000(1+0.02)^1$

$A = 81,600$

For question (II):

$A = 80000(1+0.02)^{12} - m$

$A = 101,459 - m$

Using this information, can you see what (III) is asking?


Let $a_n$ be the money owed in the $n^{th} $ month and let $r = 1.02$ be the rate at which the money owed increases.

Notice that $$a_n = ra_{n-1} \space\text{ and } \space a_0 = \$80,000 \space \text{ for } \space 0\le n \le 12$$ We can find an explicit formula from the recursion by noting the pattern (and proving it with induction if you feel the need) to get

$$a_n = a_0 r^n \space \text{ for } \space 0 \le n \le 12$$ After a year, then, $$a_{12} = a_0 r^{12} \implies a_{12} = a_0 r^{12} - M \space \text{ after paying the annual installment }$$

Now define $b_n = b_{n-1}r $ with $b_0 = a_0 r^{12} - M$

After another year and payment, we get $$b_{12} = b_0r^{12}- M = (a_0 r^{12} - M)r^{12} - M = 0$$

Solve for $M$ to get $$M = \frac {a_0r^{24}}{r^{12} + 1} $$

  • $\begingroup$ Very complicated way of looking at it, but it works. Maybe the concepts aren't so complicated, but the way you've written it looks foreign to me. $\endgroup$ – user432114 Jun 4 '17 at 2:05
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    $\begingroup$ @Dodsy: That's a fair perspective. Personally, I will do any problem using sequences that can be done that way. It typically is the most intuitive and comfortable approach for me. I also prefer to derive formulas from scratch rather than use preexisting ones in many cases (at least when first learning it). I despise how mathematics is handled in grade school, where they often just throw formulas at you unjustifiably. $\endgroup$ – infinitylord Jun 4 '17 at 2:15
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    $\begingroup$ I think your answer is more than adequate, and agree with your comment fully and completely! $\endgroup$ – user432114 Jun 4 '17 at 2:24

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