Wendie attends training sessions in preparation for a running race. In the first session, she runs a distance of 1000 metres in exactly 5 minutes. For each subsequent session, the distance run is $x\%$ more than the distance run in the previous session and the time taken is 1.5 minutes longer than the time taken for the previous session.

1. Identify the sequence as a geometric progression and find its parameters

The distance run in each session forms a geometric progression.
First term, $a = 1000$
Common ratio, $r = 1 + \frac{8}{100} = 1.08$

2. Use the formula for the $n$th term to find the distance in the 15th session

Let $u_n$ be the distance in the $n$th session.
$u_{15} = a r^{15-1}$
$u_{15} = 1000(1.08)^{14}$
$u_{15} = 2937.1938...$

3. Round the answer to 2 decimal places as required

$u_{15} \approx 2937.19 \text{ m}$

1. Identify the arithmetic progression for time and convert the target total time to minutes.

The time taken for each session forms an arithmetic progression with first term $a = 5$ minutes and common difference $d = 1.5$ minutes.
Total running time required $= 15 \text{ hours} = 15 \times 60 = 900 \text{ minutes}$.

2. Set up an inequality using the formula for the sum of an arithmetic series.

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:
$\frac{n}{2}[2(5) + (n-1)(1.5)] \ge 900$
$\frac{n}{2}[10 + 1.5n - 1.5] \ge 900$
$\frac{n}{2}[1.5n + 8.5] \ge 900$
$1.5n^2 + 8.5n - 1800 \ge 0$

3. Solve the quadratic inequality to find the minimum integer value for $n$.

Solving the equation $1.5n^2 + 8.5n - 1800 = 0$ using the quadratic formula:
$n = \frac{-8.5 \pm \sqrt{8.5^2 - 4(1.5)(-1800)}}{2(1.5)}$
$n = \frac{-8.5 \pm \sqrt{10872.25}}{3}$
$n \approx 31.92 \quad \text{or} \quad n \approx -37.59 \text{ (rejected)}$
Since $n$ must be an integer and $n \ge 31.92$, the minimum number of training sessions required is $32$.

1. Find the time taken in the $n$th session.

Let $T_n$ be the time taken in the $n$th session.
Since the time increases by $1.5$ minutes each session, it forms an arithmetic sequence with $a = 5$ and $d = 1.5$.
$T_n = 5 + (n - 1)(1.5)$
$T_n = 1.5n + 3.5$

2. Formulate the inequality for the average speed in the $n$th session.

The distance run in the $n$th session, $D_n = 1000(1.08)^{n-1}$.
Average speed in the $n$th session $= \frac{D_n}{T_n} = \frac{1000(1.08)^{n-1}}{1.5n + 3.5}$
We require the average speed to be greater than 220 m/min:
$\frac{1000(1.08)^{n-1}}{1.5n + 3.5} > 220$

3. Use a graphing calculator to find the minimum integer value of $n$.

Using the table function of a graphing calculator to evaluate $S_n = \frac{1000(1.08)^{n-1}}{1.5n + 3.5}$:
When $n = 32$, $S_{32} \approx 211.02 \text{ m/min}$
When $n = 33$, $S_{33} \approx 221.45 \text{ m/min}$
Since $221.45 > 220$, the average speed first exceeds 220 metres per minute in the 33rd session.
$n = 33$

1. Formulate the equation for the total distance run over 25 sessions.

Let the common ratio be $r = 1 + \frac{x}{100}$.
The sequence of distances forms a geometric progression with first term $a = 1000\text{ m}$.
Total distance in 25 sessions, $S_{25} = 250\text{ km} = 250000\text{ m}$.
Using the formula for the sum of a geometric series $S_n = \frac{a(r^n - 1)}{r - 1}$:
$250000 = \frac{1000(r^{25} - 1)}{r - 1}$

2. Solve the equation using a Graphing Calculator to find $r$.

Simplifying the equation by dividing both sides by 1000:
$\frac{r^{25} - 1}{r - 1} = 250$
Since the distance increases, $x > 0$ and thus $r > 1$.
Using a Graphing Calculator to solve $\frac{r^{25} - 1}{r - 1} - 250 = 0$:
$r \approx 1.160197$

3. Calculate the final value of $x$.

$1 + \frac{x}{100} = 1.160197$
$\frac{x}{100} = 0.160197$
$x = 16.0197$
$x \approx 16.02 \text{ (to 2 d.p.)}$

1. Identify the parameters of the sequence and calculate its sum to infinity.

The distances run form a geometric progression with first term $a = 2000$ and common ratio $r = 1 - 0.10 = 0.9$.
The maximum possible accumulated distance is given by the sum to infinity:
$S_\infty = \frac{a}{1 - r}$
$S_\infty = \frac{2000}{1 - 0.9} = \frac{2000}{0.1} = 20000 \text{ m}$

2. Convert the theoretical limit to kilometres and state the conclusion.

Convert the maximum distance into kilometres:
$20000 \text{ m} = 20 \text{ km}$
Since $a > 0$ and the common ratio $r = 0.9$ is positive, each session adds a strictly positive distance. For any finite number of sessions $n$, the accumulated distance $S_n$ will always be strictly less than the sum to infinity $S_\infty$.
Thus, $S_n < 20 \text{ km}$, showing she will never accumulate a total distance exceeding $20 \text{ km}$. (Shown)