# Minimum Loan Repayment and Number of Months of Loan Repayment

Jane borrows \( \$100 \ 000 \), which is to be repaid in equal monthly instalments. The interest rate is \( 6 \% \) per annum reducible, compounded monthly.

It can be shown that the amount, \( \$b_n \), owing after the \( n \) th repayment is given by the formula:

$$ b_n = 100 \ 000 r^n-x(1+r+r^2+ \cdots + r^{n-1}) $$

where \( r = 1.005 \) and \( \$ x \) is the monthly repayment.

## Part 1

The minimum monthly repayment is the amount required to repay the loan in \( 200 \) instalments. Find the minimum monthly repayment.

\( \displaystyle \begin{align} 0 &= 100 \ 000 \times 1.005^{200}-x (1.005^0+1.005^1+1.005^2+ \cdots + 1.005^{199}) \\ 0 &= 100 \ 000 \times 1.005^{200}-x \times \frac{1.005^{200}-1}{1.005-1} \\ x \times \frac{1.005^{200}-1}{0.005} &= 100 \ 000 \times 1.005^{200} \\ x &= 100 \ 000 \times 1.005^{200} \times \frac{0.005}{1.005^{200}-1} \\ &= 792.1384 \cdots \require{AMSsymbols} \\ \therefore x &= \$ 792.14 \end{align} \)

## Part 2

Jane wants to make repayments of \( \$ 1500 \) each month from the beginning of the loan. How many months will it take for her to repay the loan?

\( \displaystyle \begin{align} 100 \ 000 \times 1.005^n-1500 \times \frac{1.005^n-1}{1.005-1} &= 0 \\100 \ 000 \times 1.005^n-300 \ 000 \times (1.005^n-1) &= 0 \\100 \ 000 \times 1.005^n-300 \ 000 \times 1.005^n + 300 \ 000 &= 0 \\ -200 \ 000 \times 1.005^n &= -300 \ 000 \\ 1.005^n &= 1.5 \\ n &= \log_{1.005} 1.5 \\ &= 81.29 \cdots \\ \require{AMSsymbols} \therefore n &= 82 \text{ months} \end{align} \)

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