迦勒正在为垂直马拉松进行训练，方法是跑一栋 62 层大楼的楼梯。他决定每周训练三天，第一天训练跑 10 层。在之后的每个训练日，他跑的楼层都比前一个训练日多 2 层。

1. Identify the parameters of the arithmetic progression

Let $u_n$ be the number of floors completed on the $n^{\text{th}}$ training day.
Since he runs 2 floors more each subsequent day, the sequence is an arithmetic progression with first term $a = 10$ and common difference $d = 2$.

2. Calculate the 10th term

Using the formula $u_n = a + (n-1)d$,
$u_{10} = 10 + (10 - 1)(2)$
$u_{10} = 10 + 18 = 28$

1. Set up an equation for the number of training days $n$ required to reach 62 floors using the arithmetic progression formula.

Let $u_n$ be the number of floors completed on the $n$th training day.
$u_n = a + (n-1)d$
$10 + (n-1)(2) = 62$

2. Solve for $n$ to find the total number of training days.

$2(n-1) = 52$
$n-1 = 26$
$n = 27$

3. Determine the number of weeks $k$ by dividing the number of training days by 3.

$k = \frac{27}{3}$
$k = 9$

1. Determine the total number of training days in 20 weeks and identify the training day he reaches the cap of 62 floors.

Total training days in 20 weeks $= 20 \times 3 = 60$

Let $n$ be the training day he reaches 62 floors. Using the formula for the $n$-th term of an arithmetic progression:
$10 + (n - 1)2 = 62$
$2(n - 1) = 52$
$n - 1 = 26 \implies n = 27$

2. Calculate the total number of floors completed during the first 27 days where the daily amount is increasing.

Sum of floors for the first 27 days, $S_{27}$:
$S_{27} = \frac{27}{2}(10 + 62)$
$S_{27} = \frac{27}{2}(72) = 972$

3. Calculate the floors completed in the remaining days at the maximum rate, and find the overall total.

Remaining training days $= 60 - 27 = 33$
Floors completed in the remaining days $= 33 \times 62 = 2046$

Total floors completed in 20 weeks $= 972 + 2046 = 3018$

1. Formulate the expression for Caleb's heart rate at the start of the $n$th minute.

Let $H_n$ be his heart rate at the start of the $n$th minute.
The heart rate decreases by $10\%$ each minute, so it retains $90\%$ ($0.9$) of its previous value. This forms a geometric progression with first term $a = 180$ and common ratio $r = 0.9$.
$H_n = 180(0.9)^{n-1}$

2. Set up and solve the inequality for the heart rate to drop below 82 bpm.

We require $H_n < 82$:
$180(0.9)^{n-1} < 82$
$(0.9)^{n-1} < \frac{82}{180} = \frac{41}{90}$
Taking the natural logarithm on both sides:
$(n-1) \ln(0.9) < \ln\left(\frac{41}{90}\right)$
Since $\ln(0.9)$ is negative, dividing by it flips the inequality sign:
$n - 1 > \frac{\ln(41/90)}{\ln(0.9)}$
$n - 1 > 7.4625...$
$n > 8.4625...$

3. Determine the least integer value of $n$.

Since $n$ must be an integer, the least value of $n$ satisfying the inequality is $9$.
(Note: This can also be solved or verified using a Graphing Calculator table: $H_8 \approx 86.1 > 82$ and $H_9 \approx 77.5 < 82$, hence $n = 9$.)

1. Generate terms of the sequence iteratively using the recurrence relation to find when the heart rate drops below 90 bpm

Using the relation $H_{n+1} = 0.85H_n + 10.5$ with $H_1 = 170$, we can generate the terms of the sequence (e.g. by using a graphing calculator or computing iteratively):
$H_1 = 170$
$H_2 = 0.85(170) + 10.5 = 155$
$H_3 = 0.85(155) + 10.5 = 142.25$
$\vdots$
$H_{10} \approx 93.162$
$H_{11} \approx 89.687$

2. Identify the least value of n

Since $H_{10} \approx 93.2 > 90$ and $H_{11} \approx 89.7 < 90$, the heart rate first drops below $90\text{ bpm}$ at the $11^{\text{th}}$ minute.
The least value of $n$ is $11$.

1. Set up the equation for the steady-state heart rate as $n \to \infty$.

After resting long enough, $n \to \infty$ and the heart rate reaches a steady limit $L$. Thus, $H_{n+1} \approx H_n = L$.
$L = 0.85L + 10.5$

2. Solve the linear equation to find the resting heart rate.

$L - 0.85L = 10.5$
$0.15L = 10.5$
$L = \frac{10.5}{0.15} = 70$