Selena decides to borrow $200 000 from the bank to fund her home purchase on 1 January 2025. On the first day of each month, 0.5% interest is added to the amount owed with the first interest amount added on 1 January 2025. On the last day of each month, she makes a repayment of $x to the bank, starting from 31 January 2025.

1. Express the amount owed at the end of the 1st month.

Let $A_n$ be the amount owed at the end of month $n$.
For the first month, the interest is added before the first repayment is made:
$A_1 = 200\,000(1.005) - x$

2. Determine the amounts for subsequent months and generalise the pattern for $n$ months.

For month 2, interest is applied to $A_1$, followed by the repayment:
$A_2 = A_1(1.005) - x$
$A_2 = [200\,000(1.005) - x](1.005) - x$
$A_2 = 200\,000(1.005)^2 - x(1.005) - x$

For month 3:
$A_3 = A_2(1.005) - x$
$A_3 = 200\,000(1.005)^3 - x(1.005)^2 - x(1.005) - x$

By continuing this pattern, the amount at the end of $n$ months is:
$A_n = 200\,000(1.005)^n - x(1.005)^{n-1} - x(1.005)^{n-2} - \ldots - x$

3. Factorise out the repayment term to identify the geometric series.

$A_n = 200\,000(1.005)^n - x[1.005^{n-1} + 1.005^{n-2} + \ldots + 1.005 + 1]$
$A_n = 200\,000(1.005)^n - x[1 + 1.005 + \ldots + 1.005^{n-2} + 1.005^{n-1}]$

4. Apply the geometric progression sum formula to prove the given result.

The series inside the square brackets is a geometric series with first term $a = 1$, common ratio $r = 1.005$, and $n$ terms.
Sum $= \frac{a(r^n - 1)}{r - 1} = \frac{1(1.005^n - 1)}{1.005 - 1}$
Sum $= \frac{1.005^n - 1}{0.005} = 200(1.005^n - 1)$

Substituting this back into the expression for $A_n$:
$A_n = 200\,000(1.005)^n - x[200(1.005^n - 1)]$
$A_n = 200\,000(1.005)^n - 200x(1.005^n - 1)$ (Shown)

1. Substitute the repayment amount into the given formula and set up an inequality to find when the loan clears.

Substitute $x = 1500$ into the formula for the amount owed:
$A_n = 200\,000(1.005)^n - 200(1500)(1.005^n - 1)$
$A_n = 200\,000(1.005)^n - 300\,000(1.005^n - 1)$
$A_n = 300\,000 - 100\,000(1.005)^n$
To find when the loan is fully repaid, the amount owed must be less than or equal to zero ($A_n \le 0$):
$300\,000 - 100\,000(1.005)^n \le 0$
$(1.005)^n \ge 3$

2. Solve for the number of months $n$.

Taking the natural logarithm on both sides:
$n \ln 1.005 \ge \ln 3$
$n \ge \frac{\ln 3}{\ln 1.005}$
$n \ge 220.27$
Since repayments are made at the end of each month, the loan will be fully repaid on the 221st month ($n = 221$).

3. Determine the exact calendar date corresponding to the 221st month.

Convert 221 months into years and months:
$221 = 18 \times 12 + 5$
This represents 18 years and 5 months.
Starting from Month 1 in January 2025, 18 years later is 2043.
The 5th month of the year is May.
Because repayments are always made on the last day of the month, the date is 31 May 2043.

4. Calculate the amount of the last repayment.

The amount owed at the end of the 220th month (before the final month's interest and payment) is:
$A_{220} = 300\,000 - 100\,000(1.005)^{220} \approx 405.5637$
In the 221st month, 0.5% interest is added to this balance on the first day, and the final repayment strictly clears this amount on the last day.
$\text{Final repayment} = A_{220} \times 1.005$
$= 405.5637 \times 1.005 \approx 407.5915$
The last repayment is therefore $\char36{}407.59$.

(Alternatively, if she had paid $\char36{}1500$, the balance would be $A_{221} = 300\,000 - 100\,000(1.005)^{221} = -1092.41$, representing an overpayment. Her required payment is $1500 - 1092.41 = \char36{}407.59$.)

1. Determine the number of months in 10 years and substitute into the given formula, setting the amount owed to zero.

$n = 10 \times 12 = 120$

For the loan to be fully repaid, the amount owed at the end of 120 months must be $0$:
$200\,000(1.005)^{120} - 200x(1.005^{120} - 1) = 0$

2. Solve for the monthly repayment $x$.

$200x(1.005^{120} - 1) = 200\,000(1.005)^{120}$
$x = \frac{200\,000(1.005)^{120}}{200(1.005^{120} - 1)}$
$x = \frac{1000(1.005)^{120}}{1.005^{120} - 1}$
$x = 2220.41001...$
$x \approx 2220.41$