A person borrows Rs $640000$ at a nominal interest rate of $10\%$ per annum, compounded continuously. The borrower makes payments continuously at a constant rate of Rs $k$ per year such that the loan is paid off in exactly $3$ years. Determine $k,$ and the total interest paid.
I am not sure but I think, if the borrower is supposed to pay back the interest in 3 years then the final amount to be paid back in three years is calculated and then he can pay that same amount over time. Or will the fact that he pays continously as well affect the interest too? Here is how I solved the exercise; is it correct?
Principal amount $P = \text{Rs.} 640,000$, rate of interest is $r = 10\%$ per year, and time is $t = 3$ years. If $n$ is the number of times that the interest is compounded in one time period, which is $1$ year in this case, then the total amount to be paid is $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$
If interest is compounded continuously, then we can say that $n \to \infty$, so, $$A = \lim_{n \to \infty} P\left(1 + \frac{r}{n}\right)^{nt}$$ $$\implies A = P\cdot e^{rt}$$ And if the constant annual payment rate is $k$ and the loan is to be paid back in 3 years, then we must have, $$P\cdot e^{rt} = kt$$ $$\implies k = \frac{P\cdot e^{rt}}{t}$$ $$\implies k = \frac{\text{Rs.} \ 640,000\cdot e^{0.1 \times 3}}{3\ \text{years}}$$ $$\implies k \approx \text{Rs.}\ 287969.879 \ \text{per year}$$ And the net interest to be paid is $A - P = P(e^{0.3} - 1) \approx \text{Rs.}\ 223909.637$
