A loan has to be repaid over 3n years, through annual repayment installments of 5000 each at an annual interest rate of 10%. Sum of the Interests paid in the (n+1) and that in the 2n+1 year is 5000.

calculate the number of years over which the loan is to be repaid.

I normally use the present value or future value formula, unsure how to use those formulas in this situation, as I dont have a value for the debt.

any help appreciated.


If $P(n)$ is the amount due at the end of year $n$, and assuming payment at the end of the year you have

$$ P(n+1) = P(n) + 0.1 P(n) -5000 $$ If the payment is made at the start of the year than use $$ P(n+1) = 1.1 (P(n)-5000 $$

Also total interest paid up to and including year n is $$ I(n+1) = I(n) + 0.1 P(n) $$

$P(0)$ is the initial principal and $I(0)=0$ and $P(3n)=0$.

First show that $$P(n) = A (1.1)^n + B $$ and $$I(n) = C \cdot n + D (1.1)^n + E$$

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    $\begingroup$ There's a lot nicer way of solving this using amortization, see my answer. $\endgroup$ – Tyler Jan 8 '14 at 20:27

If we let $I_k$ and $R_k$ be the interest and principal, respectively, of the $k$th payment, then we know that $I_k + R_k$ = 5000. We can easily calculate (using an amortization table) that $R_k = Pv^{N - k+1}$ where $P$ is the annual payment and $N$ is the term of the loan.

Thus we know that:

$5000 = I_{2n+1} + I_{n+1}= 5000(1 - v^{3n - (2n+1) - 1}) + 5000(1 - v^{3n - (n+1) - 1})$


$1 = 2 - v^{n} - v^{2n}$.

Then just solve the quadratic.


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