踏频，即骑行时的脚踏速度，以每分钟脚踏旋转圈数（RPM）为单位进行测量。例如，60 RPM 的踏频意味着一个脚踏在 1 分钟内完成 60 次完整旋转。动感单车上脚踏的每一次完整旋转相当于骑行了大约 0.006 km 的距离。 Lisa 报名参加了一个初级动感单车课程。训练计划要求她在第 1 分钟内持续以 30 RPM 的踏频骑行，并在随后的每一分钟内逐渐将踏频提高 2.5 RPM。 Celina 报名参加了一个中级动感单车课程。训练计划要求她在第 1 分钟内持续以 25 RPM 的踏频骑行，并在随后的每一分钟内将踏频提高 15%。

1. Set up an equation for the $n$-th term of the arithmetic progression.

Lisa's cadence forms an arithmetic progression with first term $a = 30$ and common difference $d = 2.5$.
Let $n$ be the minute at which she cycles at 120 RPM.
$u_n = a + (n-1)d$
$120 = 30 + (n-1)(2.5)$

2. Solve for $n$.

$120 - 30 = 2.5(n-1)$
$90 = 2.5(n-1)$
$n-1 = \frac{90}{2.5}$
$n-1 = 36$
$n = 37$

1. Find the total number of pedal revolutions in the first 15 minutes using the sum of an arithmetic progression.

$S_{15} = \frac{15}{2} \left[2(30) + (15 - 1)(2.5)\right]$
$S_{15} = 7.5 \left[60 + 35\right]$
$S_{15} = 712.5 \text{ revolutions}$

2. Calculate the total distance cycled in kilometers.

$\text{Total distance} = 712.5 \times 0.006$
$\text{Total distance} = 4.275 \text{ km}$

3. Convert the time to hours and calculate the average speed.

$\text{Time in hours} = \frac{15}{60} = 0.25 \text{ h}$
$\text{Average speed} = \frac{4.275}{0.25}$
$\text{Average speed} = 17.1 \text{ km/h}$

1. Formulate an inequality for the cadence at the $n$-th minute using the formula for the $n$-th term of a geometric sequence.

$25(1.15)^{n-1} > 120$

2. Solve the exponential inequality using logarithms to find $n$.

$1.15^{n-1} > 4.8$
$(n-1)\ln(1.15) > \ln(4.8)$
$n - 1 > \frac{\ln(4.8)}{\ln(1.15)}$
$n - 1 > 11.223$
$n > 12.223$

3. Identify the exact minute from the calculated value of $n$.

Since $n$ must be an integer representing the minute, $n = 13$.

1. Express the total distance cycled by Lisa in $n$ minutes

Lisa's cadence follows an arithmetic progression with first term $a = 30$ and common difference $d = 2.5$.
Total revolutions in $n$ minutes, $S_L = \frac{n}{2}[2(30) + (n-1)(2.5)]$
Total distance, $D_L = 0.006 \times \frac{n}{2}[60 + 2.5(n-1)]$

2. Express the total distance cycled by Celina in $n$ minutes

Celina's cadence follows a geometric progression with first term $a = 25$ and common ratio $r = 1.15$.
Total revolutions in $n$ minutes, $S_C = \frac{25(1.15^n - 1)}{1.15 - 1}$
Total distance, $D_C = 0.006 \times \frac{25(1.15^n - 1)}{0.15}$

3. Find the smallest integer $n$ for which Celina's distance exceeds Lisa's distance

We require $D_C > D_L$.
Comparing the values of $D_C$ and $D_L$ for integer values of $n$:
When $n = 5$, $D_C \approx 1.011 \text{ km} < D_L = 1.050 \text{ km}$
When $n = 6$, $D_C \approx 1.313 \text{ km} > D_L = 1.305 \text{ km}$
Since Celina's total distance cycled first exceeds Lisa's total distance at $n=6$, this occurs at the 6th minute.