Caleb正在为一项垂直马拉松比赛进行训练，方法是爬一栋62层高的大楼的楼梯。他决定每周训练三天，第一天训练时爬10层。在之后的每个训练日，他都比前一个训练日多爬2层。 一旦他能在一个训练日内完成62层，他就被认为为垂直马拉松做好了准备。然后，在随后的训练日里，他将继续每天完成62层。 训练结束后，Caleb的心率在每个第 $n$ 分钟开始时降低10%。当 $n = 1$ 时，他在第一分钟开始时的心率为每分钟180次（bpm）。 Caleb的朋友Alex也记录了自己训练后的心率。他在第一分钟开始时的心率为170 bpm，他的心率可以用递推关系式 $H_{n+1} = 0.85 H_n + 10.5$ 来建模，其中 $H_n$ 是他训练后第 $n$ 分钟开始时的心率。

1. Identify the parameters of the arithmetic progression

Let $u_n$ be the number of floors completed on the $n$-th training day.
The sequence of floors forms an arithmetic progression where the first term $a = 10$ and the common difference $d = 2$.

2. Calculate the number of floors on the 10th training day

$u_{10} = a + (10 - 1)d$
$u_{10} = 10 + 9(2)$
$u_{10} = 28$

1. Determine the training day on which Caleb completes 62 floors using the formula for the $n$th term of an arithmetic progression.

$u_n = a + (n-1)d$
$62 = 10 + (n-1)2$
$52 = 2(n-1)$
$n - 1 = 26 \implies n = 27$

2. Calculate the number of weeks $k$ needed to complete 27 training days.

Since he trains 3 days a week:
$k = \frac{27}{3} = 9$

1. Find the training day on which Caleb first completes 62 floors.

Let $n$ be the number of training days to reach 62 floors. Using the formula for the $n$th term of an arithmetic progression $u_n = a + (n - 1)d$:
$10 + (n - 1)(2) = 62$
$2(n - 1) = 52$
$n - 1 = 26$
$n = 27$

2. Calculate the total number of floors completed in the first 27 training days.

Using the sum formula for an arithmetic series $S_n = \frac{n}{2}(a + l)$:
$S_{27} = \frac{27}{2}(10 + 62)$
$S_{27} = \frac{27}{2}(72) = 972$

3. Calculate the overall total number of floors completed at the end of 20 weeks.

Total training days in 20 weeks = $20 \times 3 = 60$ days.
Remaining training days where he completes 62 floors = $60 - 27 = 33$ days.
Overall total floors completed = $972 + (33 \times 62)$
$= 972 + 2046 = 3018$

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

Let $C_n$ be Caleb's heart rate at the start of the $n$th minute.
Since it decreases by $10\%$ each minute, $C_n$ forms a geometric sequence with first term $a = 180$ and common ratio $r = 1 - 0.10 = 0.9$.
$C_n = 180(0.9)^{n-1}$

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

$180(0.9)^{n-1} < 82$
$0.9^{n-1} < \frac{82}{180}$
$0.9^{n-1} < \frac{41}{90}$

3. Solve the inequality for $n$ using logarithms to find the least integer value.

Taking the natural logarithm on both sides:
$(n-1)\ln(0.9) < \ln\left(\frac{41}{90}\right)$
Since $\ln(0.9) < 0$, the inequality sign flips when dividing:
$n-1 > \frac{\ln(41/90)}{\ln(0.9)}$
$n-1 > 7.462$
$n > 8.462$
The least integer value of $n$ is $9$.

1. Generate the sequence using a graphing calculator to find terms around the threshold of 90 bpm.

Using a graphing calculator to generate the sequence $H_{n+1} = 0.85H_n + 10.5$ with $H_1 = 170$:
When $n = 10$, $H_{10} \approx 93.16 > 90$
When $n = 11$, $H_{11} \approx 89.69 < 90$

2. Identify the least value of $n$.

Since $H_{11}$ is the first term that falls below $90$ bpm, the least value of $n$ is $11$.

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

After resting long enough, $n \to \infty$ and the heart rate will reach a steady value $L$.
As $H_{n+1} \to L$ and $H_n \to L$, substitute $L$ into Alex's recurrence relation:
$L = 0.85L + 10.5$

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

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