一支自行车队计划在 10 天内从新加坡骑行到槟城并返回。从新加坡到槟城往返的路线距离为 1600 km。 艾利克斯（Alex）建议车队第一天骑行 120 km，之后每天增加 8 km 的骑行距离。证明车队无法在 10 天内完成该路线。 贝儿（Belle）建议改变第一天的骑行距离，以便车队在第十天到达新加坡。已知之后每天增加的骑行距离仍为 8 km，且最后一天的骑行距离为 196 km，求第一天的骑行距离。 2025年1月2日，糖糖（Candy）在一个银行账户中存入 $8000，该账户在每年年底提供 1\% 的年复利。此后每年的年初都会取出$1200。 求第三年年初糖糖银行账户中的金额。 证明第 $(n+1)$ 年年初糖糖银行账户中的金额为 $S\left[120000-112000(1.01)^n\right]$。 糖糖的银行账户在第几年会首次少于 $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At the end of Year 1, $1\%$ interest is added, and at the beginning of Year 2, $\char36{}1200$ is withdrawn: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 $2025 + 6 = 2031$.