Sarah, a Singaporean Citizen, started a 4-year undergraduate Computing course at the National University of Singapore (NUS) in August 2021. The annual tuition fee, after the MOE Tuition Grant, is $8,250. Sarah took a Tuition Fee Loan (TFL) covering $90\%$ of her tuition fees. This loan is interest-free during her course of study. Sarah decided to set aside $x$ per month from August 2021. This amount is increased by $2 each subsequent month. (Sarah set aside $x$ in August 2021, $(x+2)$ in September 2021, ...). With the amount set aside, she repaid one-fifth of her total TFL at the end of July 2025. Upon Sarah's graduation, the TFL's interest rate becomes effective at a fixed rate of $0.4\%$ per month, calculated on the outstanding balance at the end of the month. The interest is first charged on 31st August 2025. Sarah plans to make monthly repayments of $y$ to the bank on the 1st of every month, starting from 1st September 2025.

1. Calculate the total loan amount and the target amount to be repaid.

$\text{Total tuition fee for 4 years} = \char36{}8250 \times 4 = \char36{}33000$

$\text{Total TFL} = 90\% \text{ of } \char36{}33000 = 0.9 \times \char36{}33000 = \char36{}29700$

$\text{Amount repaid} = \frac{1}{5} \times \char36{}29700 = \char36{}5940$

2. Express the total amount set aside over the 4 years in terms of $x$ using the arithmetic series sum formula.

The duration from August 2021 to July 2025 is exactly 4 years, which is $48$ months.

The amounts set aside form an arithmetic progression with first term $a = x$, common difference $d = 2$, and number of terms $n = 48$.

$S_{48} = \frac{n}{2} \left[ 2a + (n - 1)d \right]$
$S_{48} = \frac{48}{2} \left[ 2x + (48 - 1)(2) \right]$
$S_{48} = 24(2x + 94)$
$S_{48} = 48x + 2256$

3. Formulate an equation by equating the total savings to the amount repaid, and solve for $x$.

$48x + 2256 = 5940$
$48x = 3684$
$x = 76.75$

1. Calculate the total Tuition Fee Loan (TFL) and the outstanding loan principal before interest begins.

$\text{Total tuition fees} = 4 \times 8250 = 33000$
$\text{Total TFL taken} = 0.9 \times 33000 = 29700$

Since Sarah repaid $\frac{1}{5}$ of her total TFL at the end of July 2025, the outstanding amount is:
$\text{Outstanding loan (end July 2025)} = \frac{4}{5} \times 29700 = 23760$

2. Apply the first interest charge and subtract the first monthly repayment.

Interest is first charged on 31st August 2025 at $0.4\%$ per month.
$\text{Balance on 31st August 2025} = 23760 \times \left(1 + \frac{0.4}{100}\right)$
$= 23760 \times 1.004 = 23855.04$

After her first monthly repayment of $y$ on 1st September 2025:
$\text{Outstanding loan} = 23855.04 - y$

1. Calculate the outstanding loan for the first few months to establish a pattern.

2. Generalise the outstanding loan expression for $n$ months.

3. Apply the sum formula for a finite geometric series and substitute to show the final expression.

1. Set up the equation for the loan to be paid off in 3 years.

Since the loan is paid off in 3 years, the number of monthly repayments is $n = 3 \times 12 = 36$.
Equating the expression from part (c) to zero:
$23760(1.004)^{36} - 250y(1.004^{36} - 1) = 0$

2. Solve for $y$.

$250y(1.004^{36} - 1) = 23760(1.004)^{36}$
$250y \approx \frac{27432.0189}{0.154546}$
$y = \frac{23760(1.004)^{36}}{250(1.004^{36} - 1)}$
$y \approx 710.0007$
$y = 710.00 \text{ (to 2 d.p. for money)}$

1. Calculate the outstanding loan after 24 months of $500 repayments.

Using the formula from (c) with $y = 500$ and $n = 24$:
$\text{Outstanding loan} = 23760(1.004)^{24} - 250(500)(1.004^{24} - 1)$
$= 13576.5154$

2. Set up an equation to find the additional number of months $m$ required to clear the loan with $1000 repayments.

$13576.5154(1.004)^m - 250(1000)(1.004^m - 1) = 0$
$13576.5154(1.004)^m - 250000(1.004^m) + 250000 = 0$
$236423.4846(1.004)^m = 250000$
$(1.004)^m = \frac{250000}{236423.4846} \approx 1.05742$
$m = \frac{\ln 1.05742}{\ln 1.004} \approx 13.987$
Since $m$ must be a whole number of repayments, 14 additional months are needed.

3. Determine the calendar month and year when the loan is cleared.

Total number of months $= 24 + 14 = 38$.
The 1st repayment is in September 2025. The 38th repayment occurs 37 months later.
37 months is 3 years and 1 month.
September 2025 $+ 3$ years $=$ September 2028.
September 2028 $+ 1$ month $=$ October 2028.

4. Calculate the amount of the final repayment in the 38th month.

The final repayment is made in the 14th additional month (month 38 overall). First, find the outstanding loan after 13 additional payments (the 37th payment overall):
$A_{37} = 13576.5154(1.004)^{13} - 250000(1.004^{13} - 1)$
$A_{37} \approx 981.6602$
The final repayment covers this remaining balance plus 1 month of interest:
$\text{Final repayment} = 981.6602 \times 1.004 \approx 985.59$