A team of cyclists plans to ride from Singapore to Penang and back over 10 days. The distance for the route taken from Singapore to Penang and back is 1600 km. Alex suggests that the team cycle 120 km for the first day and increase the distance covered by 8 km each subsequent day. Show that the team will not be able to complete the route in 10 days. Belle suggests changing the distance covered on the first day so that the team will reach Singapore on the tenth day. Given that the increment of distance covered for each subsequent day remains at 8 km, and the distance covered on the last day is 196 km, find the distance covered on the first day. On 2 January 2025, Candy saved $8000 in a bank account which gives a compound interest of 1\% per annum at the end of each year. Withdrawals of$1200 are made at the beginning of each subsequent year. Find the amount in Candy's bank account at the beginning of the third year. Show that the amount in Candy's bank account at the beginning of the $(n+1)$th year is $S\left[120000-112000(1.01)^n\right]$. In which year would Candy's bank account first have less than $1200?

1. Identify the parameters of the arithmetic progression representing the distance cycled.

Let the distance cycled on the first day be $a$ and the daily increase be $d$.
$a = 120$
$d = 8$

2. Calculate the total distance covered in 10 days using the sum formula for an arithmetic progression.

$S_n = \frac{n}{2} [2a + (n-1)d]$
$S_{10} = \frac{10}{2} \left[ 2(120) + (10 - 1)(8) \right]$
$S_{10} = 5 \left[ 240 + 72 \right]$
$S_{10} = 1560$

3. Compare the total distance with the required distance to draw the conclusion.

The total distance covered in $10$ days is $1560 \text{ km}$.
Since $1560 < 1600$, the team will not be able to complete the $1600 \text{ km}$ route in $10$ days. (Shown)

1. Set up an equation using the formula for the $n$-th term of an arithmetic progression, given the distance on the 10th day.

$u_n = a + (n-1)d$
$u_{10} = a + (10 - 1)(8)$
$196 = a + 9(8)$

2. Solve for $a$, the distance covered on the first day.

$a + 72 = 196$
$a = 124$

1. Calculate the amount in the account at the beginning of the second year.

Amount at the end of the 1st year $= 8000 imes 1.01 = 8080$

Amount at the beginning of the 2nd year $= 8080 - 1200 = 6880$

2. Calculate the amount in the account at the beginning of the third year.

Amount at the end of the 2nd year $= 6880 imes 1.01 = 6948.80$

Amount at the beginning of the 3rd year $= 6948.80 - 1200 = 5748.80$

1. Determine the amount at the beginning of the first few years to establish the recurrence pattern.

Let $U_k$ be the amount in the account at the beginning of the $k$-th year. We are given $U_1 = 8000$.
At the end of Year 1, $1\%$ interest is added, and at the beginning of Year 2, $\char36{}1200$ is withdrawn:
$U_2 = 8000(1.01) - 1200$
$U_3 = [8000(1.01) - 1200](1.01) - 1200$
$U_3 = 8000(1.01)^2 - 1200(1.01) - 1200$
By continuing this pattern, the amount at the beginning of the $(n+1)$-th year is:
$U_{n+1} = 8000(1.01)^n - 1200(1.01)^{n-1} - 1200(1.01)^{n-2} - \dots - 1200$

2. Factor out the common withdrawal amount to reveal a geometric series.

$U_{n+1} = 8000(1.01)^n - 1200 \left[ (1.01)^{n-1} + (1.01)^{n-2} + \dots + 1.01 + 1 \right]$
Reversing the terms in the square bracket gives a standard geometric series:
$U_{n+1} = 8000(1.01)^n - 1200 \left[ 1 + 1.01 + 1.01^2 + \dots + (1.01)^{n-1} \right]$

3. Evaluate the sum of the finite geometric series.

The series inside the brackets is a geometric progression with first term $a = 1$, common ratio $r = 1.01$, and $n$ terms.
Using the sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$:
$S_n = \frac{1 \cdot ((1.01)^n - 1)}{1.01 - 1} = \frac{(1.01)^n - 1}{0.01} = 100((1.01)^n - 1)$

4. Substitute the evaluated sum back and simplify to obtain the final expression.

$U_{n+1} = 8000(1.01)^n - 1200 \left[ 100((1.01)^n - 1) \right]$
$U_{n+1} = 8000(1.01)^n - 120000((1.01)^n - 1)$
$U_{n+1} = 8000(1.01)^n - 120000(1.01)^n + 120000$
$U_{n+1} = 120000 - 112000(1.01)^n$
This matches the target expression $\char36{}\left[ 120000 - 112000(1.01)^n \right]$ (noting that the $S$ in the question is a typographical error for $\char36{}$).

1. Set up an inequality using the expression from part (b)(ii).

Using the expression given for the amount at the beginning of the $(n+1)$th year, we want to find when this amount first drops below $\char36{}1200$:
$120000 - 112000(1.01)^n < 1200$
$112000(1.01)^n > 120000 - 1200$
$112000(1.01)^n > 118800$
$(1.01)^n > \frac{118800}{112000}$
$(1.01)^n > \frac{297}{280}$

2. Solve the exponential inequality for $n$.

Taking the natural logarithm on both sides:
$\ln(1.01)^n > \ln\left(\frac{297}{280}\right)$
$n \ln(1.01) > \ln\left(\frac{297}{280}\right)$
$n > \frac{\ln(297/280)}{\ln(1.01)}$
$n > 5.9188\dots$

3. Determine the first calendar year the condition is met.

Since $n$ must be an integer, the smallest integer satisfying the inequality is $n = 6$.
The expression corresponds to the beginning of the $(n+1)$th year.
For $n = 0$, the $1$st year is $2025$.
For $n = 6$, it corresponds to the $(6+1)$th year, which is the $7$th year.
The calendar year is $2025 + 6 = 2031$.