学校的步数挑战赛是体育部发起的一项活动，旨在鼓励学生过上更积极健康的生活方式。下表显示了相应的每日步数以及可获得的健康积分。 学生 A 第一天走了 4500 步。此后每天，他的步数比前一天增加 350 步。

1. Formulate an inequality for the daily steps to reach the 15 Health Points category.

Student A's daily steps form an arithmetic progression with first term $a = 4500$ and common difference $d = 350$.
To earn 15 Health Points, the step count must be at least 7500.
$u_n = a + (n-1)d \ge 7500$
$4500 + (n-1)350 \ge 7500$

2. Solve the inequality to find the first day $n$ this occurs.

$350(n-1) \ge 3000$
$n - 1 \ge \frac{3000}{350}$
$n - 1 \ge 8.571$
$n \ge 9.571$
Since $n$ must be an integer, the first such day is $n = 10$.

1. Find the day when Student A first earns 5 points and the day they first earn 25 points.

Let $n$ be the day number and $u_n$ be the number of steps on day $n$.
$u_n = 4500 + 350(n-1)$
For $\ge 5000$ steps (5 points tier):
$4500 + 350(n-1) \ge 5000 \implies 350(n-1) \ge 500 \implies n \ge 2.43$
Student A starts earning 5 points on Day 3.

For $\ge 10000$ steps (25 points tier):
$4500 + 350(n-1) \ge 10000 \implies 350(n-1) \ge 5500 \implies n \ge 16.71$
Student A starts earning 25 points on Day 17.

(Note: As found in part (a)(i), Student A starts earning 15 points on Day 10).
(Also, $u_1 = 4500$ and $u_2 = 4850$, both of which are $< 5000$ and thus earn 0 points)..

2. Determine the number of days spent in each points tier from Day 1 to Day 20.

By grouping the 20 days into their respective tiers:

• 0 points tier ($< 5000$ steps): Days 1 to 2 ($2$ days)
• 5 points tier ($5000 \le u_n \le 7499$): Days 3 to 9 ($9 - 3 + 1 = 7$ days)
• 15 points tier ($7500 \le u_n \le 9999$): Days 10 to 16 ($16 - 10 + 1 = 7$ days)
• 25 points tier ($\ge 10000$ steps): Days 17 to 20 ($20 - 17 + 1 = 4$ days)

3. Calculate the total health points accumulated by the end of the 20th day.

Multiply the number of days in each tier by the respective points and sum them up:
$\text{Total Points} = (2 \times 0) + (7 \times 5) + (7 \times 15) + (4 \times 25)$
$= 0 + 35 + 105 + 100$
$= 240$

1. Set up an inequality for the total number of steps using the geometric series sum formula.

Let the common ratio be $r = 1 + \frac{x}{100}$.
The total number of steps in 20 days is given by the sum of the first 20 terms of a geometric progression with first term $a = 7500$.
$S_{20} = \frac{7500(r^{20} - 1)}{r - 1} \ge 261500$

2. Simplify the inequality.

$\frac{r^{20} - 1}{r - 1} \ge \frac{261500}{7500}$
$\frac{r^{20} - 1}{r - 1} \ge \frac{523}{15}$

3. Solve for $x$ using a graphing calculator and round to the required precision.

Using a graphing calculator to solve $\frac{r^{20} - 1}{r - 1} = \frac{523}{15}$, we obtain:
$r \approx 1.054996$
Substitute $r = 1 + \frac{x}{100}$ back into the inequality:
$1 + \frac{x}{100} \ge 1.054996$
$x \ge 5.4996$
Rounding to 2 decimal places, the minimum value of $x$ is $5.50$.

1. Formulate the sequence for the daily step count.

Let the number of steps on day $n$ be $u_n$.
The step count forms a geometric progression with the first term $a = 7500$ and a common ratio $r = 1 + \frac{3.80}{100} = 1.038$.
$u_n = 7500(1.038)^{n-1}$

2. Find the first day when the step count reaches at least 10000.

We need $u_n \ge 10000$:
$7500(1.038)^{n-1} \ge 10000$
$(1.038)^{n-1} \ge \frac{10000}{7500} = \frac{4}{3}$
Taking the natural logarithm on both sides:
$(n-1) \ln(1.038) \ge \ln\left(\frac{4}{3}\right)$
$n - 1 \ge \frac{\ln(4/3)}{\ln 1.038} \approx 7.7135$
$n \ge 8.7135$
Thus, the step count reaches 10000 or more starting from Day 9.

3. Calculate the total health points over 20 days.

From Day 1 to Day 8 (total $8$ days), the step count is between $7500$ and $9999$, earning $15$ points per day.
From Day 9 to Day 20 (total $20 - 9 + 1 = 12$ days), the step count is $\ge 10000$, earning the maximum of $25$ points per day.
$\text{Total points} = 8(15) + 12(25)$
$\text{Total points} = 120 + 300 = 420$

1. Identify the first term and the common ratio of the geometric progression.

First term, $a = 2$
Common ratio, $r = 1 - 0.09 = 0.91$

2. Calculate the sum to infinity to find the maximum possible accumulated value and conclude.

Maximum possible total value $= S_\infty = \frac{a}{1 - r}$
$S_\infty = \frac{2}{1 - 0.91} = \frac{2}{0.09} = 22\frac{2}{9}$
Since $22\frac{2}{9} \approx 22.22$, which is strictly less than $23$, the accumulated value will not exceed $\char36{}23$. (Shown)