Jeremy plans to invest $100 000 at the beginning of year 2024 into a current account which pays compound interest at a rate of 3.2% per annum. The interest is paid on the last day of each year based on the amount in the account. He also puts a further $X into the current account on the first day of each subsequent year.

1. Calculate the amount in the account at the end of 2024 (the first year).

$\text{Amount at the end of 2024} = 100000 \times \left(1 + \frac{3.2}{100}\right) = 100000 \times 1.032 = 103200$

2. Form an expression for the amount at the end of 2025 by adding $X$ and applying interest again.

$\text{Amount at the beginning of 2025} = 103200 + X$
$\text{Amount at the end of 2025} = (103200 + X) \times 1.032$
$= 106502.4 + 1.032X$

1. Formulate the pattern for the total amount at the end of the $n$th year.

Let $A_n$ be the amount at the end of year $n$.
End of Year 1 (2024): $A_1 = 100000(1.032)$
End of Year 2 (2025): $A_2 = 100000(1.032)^2 + X(1.032)$
End of Year 3 (2026): $A_3 = 100000(1.032)^3 + X(1.032)^2 + X(1.032)$
Following this pattern, by the end of the $n$th year:
$A_n = 100000(1.032)^n + X(1.032^{n-1}) + X(1.032^{n-2}) + \dots + X(1.032)$

2. Identify the geometric series formed by the $X$ contributions and state its parameters.

The sequence of terms involving $X$ forms a geometric series:
$X(1.032) + X(1.032^2) + \dots + X(1.032^{n-1})$
First term, $a = 1.032X$
Common ratio, $r = 1.032$
Number of terms $N = n - 1$

3. Apply the sum formula for a geometric progression and simplify the final expression.

Using the sum formula $S_N = \frac{a(r^N-1)}{r-1}$:
Sum of $X$ terms $= \frac{1.032X(1.032^{n-1} - 1)}{1.032 - 1}$
$= \frac{1.032X}{0.032}(1.032^{n-1} - 1)$
$= 32.25X(1.032^{n-1} - 1)$

Combining this with the initial investment's compounded amount gives the total amount $A_n$:
$A_n = 100000(1.032)^n + 32.25X(1.032^{n-1} - 1)$

1. Find an expression for the amount at the end of the $n$-th year using $X = 8000$.

Using the expression for the sum from part (b), where the interest multiplier is $1.032$:
$S_n = 100\,000(1.032)^n + \frac{8000(1.032)(1.032^{n-1} - 1)}{1.032 - 1}$
$S_n = 100\,000(1.032)^n + 258\,000(1.032^{n-1} - 1)$
$S_n = 100\,000(1.032)^n + 250\,000(1.032)^n - 258\,000$
$S_n = 350\,000(1.032)^n - 258\,000$

2. Solve for the year when the end-of-year balance reaches $170,000.

Set $S_n \ge 170\,000$:
$350\,000(1.032)^n - 258\,000 \ge 170\,000$
$350\,000(1.032)^n \ge 428\,000$
$1.032^n \ge \frac{428}{350} \approx 1.22286$
$n \ge \frac{\ln(1.22286)}{\ln(1.032)} \approx 6.386$
This means at the end of year 6, the amount is less than $\char36{}170\,000$.

3. Check the amount on the first day of the following year to conclude the exact year and day.

Amount at the end of year 6 ($n=6$):
$S_6 = 350\,000(1.032)^6 - 258\,000 \approx 164\,810.96$

Amount on the first day of year 7 (after depositing the additional $\char36{}8000$):
$164\,810.96 + 8000 = 172\,810.96$

Since $\char36{}172\,810.96 \ge \char36{}170\,000$, Jeremy will have enough money on the first day of year 7.
Since year 1 is 2024, year 7 is $2024 + 7 - 1 = 2030$.

1. Calculate the total simple interest over $m$ years

$\text{Total Interest} = 170000 \times 0.028 \times m = 4760m$

2. Calculate the annual instalment

$\text{Total Repayment Amount} = 170000 + 4760m$

$\text{Annual instalment} = \frac{170000 + 4760m}{m} = \frac{1}{m}(170000 + 4760m) \quad \text{(Shown)}$

1. Calculate the annual instalment for the car loan and the annual deduction from the current account.

Substitute $m = 10$ into the formula from part (d):
$\text{Annual instalment} = \frac{1}{10}(170000 + 4760(10)) = \frac{217600}{10} = \$21760$

Since Jeremy uses $\char36{}8000$ to partially pay this instalment, the amount to be deducted from his current account at the end of each year is:
$\text{Annual deduction} = 21760 - 8000 = \$13760$

2. Formulate the future value of the current account after the 10-year loan period.

Assuming the loan is taken at the start of 2025, the balance at the end of 2024 before deductions is:
$\text{Balance at end of 2024} = 100000 \times 1.032 = \$103200$

For the next 10 years (end of 2025 to end of 2034), this initial sum accrues compound interest, while $\char36{}13760$ is deducted at the end of each year.
Let $A$ be the final balance at the end of 2034:
$A = 103200(1.032)^{10} - \left[ 13760(1.032)^9 + 13760(1.032)^8 + \dots + 13760(1.032)^0 \right]$

The deductions form a geometric progression with $a = 13760$, $r = 1.032$, and $n = 10$. Using the sum of a geometric series:
$A = 103200(1.032)^{10} - 13760 \left( \frac{1.032^{10} - 1}{1.032 - 1} \right)$

3. Evaluate the balance to justify whether the arrangement is possible.

$A = 141409.19 - 13760 \left( \frac{1.370244 - 1}{0.032} \right)$
$A = 141409.19 - 159204.98$
$A = -17795.79$

Since the projected final balance is negative (the accumulated value of the deductions exceeds the accumulated value of the initial deposit), the current account will not have sufficient funds to sustain the deductions for the full 10 years. Therefore, this arrangement is not possible.