Adam 和 Gregory 报名参加了马拉松比赛。为了准备这次马拉松，Adam 和 Gregory 各自制定了一个为期 15 周的个性化训练计划。Adam 在第 1 周的第一天跑了 2.4 km，在随后每周的第一天，所跑的距离比前一周增加 20%。Gregory 在第 1 周的第一天也跑了 2.4 km，但在随后每周的第一天，所跑的距离增加 $d$ km，其中 $d$ 是一个常数。假设 Adam 和 Gregory 仅在每周的第一天跑步。

1. Identify the sequence of Gregory's distances as an arithmetic progression

Gregory's weekly distances form an arithmetic progression where the first term $a = 2.4$, the common difference is $d$, and the number of terms $n = 15$.

2. Apply the sum formula for an arithmetic progression and simplify

The total distance is the sum of the 15 terms, $S_{15}$.
$S_n = \frac{n}{2} [2a + (n-1)d]$
$S_{15} = \frac{15}{2} [2(2.4) + (15-1)d]$
$S_{15} = \frac{15}{2} [4.8 + 14d]$
$S_{15} = 15 [2.4 + 7d] = 36 + 105d$

1. Formulate the total distance covered by Adam as the sum of a geometric progression.

Adam's weekly distances form a geometric progression with first term $a = 2.4$ and common ratio $r = 1 + 0.2 = 1.2$.

Total distance in 15 weeks, $S_{15} = \frac{2.4(1.2^{15} - 1)}{1.2 - 1}$

2. Calculate the total distance and conclude whether the target is achieved.

$S_{15} = \frac{2.4(15.407 - 1)}{0.2} \approx 172.88 \text{ km}$ (to 5 s.f.)

Since $172.88 \text{ km} > 170 \text{ km}$, Adam can achieve his target.

1. Find the expressions for the distance covered by Adam and Gregory on the first day of the 15th week.

Adam's distance (Geometric Progression with $a=2.4$, $r=1.2$):
$A_{15} = 2.4 \times 1.2^{15-1} = 2.4 \times 1.2^{14}$

Gregory's distance (Arithmetic Progression with $a=2.4$, common difference $d$):
$G_{15} = 2.4 + (15-1)d = 2.4 + 14d$

2. Formulate an inequality based on the condition that Adam covers a longer distance than Gregory.

$A_{15} > G_{15}$
$2.4 \times 1.2^{14} > 2.4 + 14d$
$30.81404 > 2.4 + 14d$
$14d < 28.41404$
$d < 2.02957$

3. Determine the maximum value of $d$ correct to 2 decimal places.

Since $d$ must be strictly less than $2.02957...$ and needs to be given to 2 decimal places, the maximum possible 2-decimal-place value is:
$d = 2.02$

4. Show that the difference in distance covered is 0.134 km correct to 3 significant figures.

Substitute $d = 2.02$ into Gregory's distance:
$G_{15} = 2.4 + 14(2.02) = 30.68$

Difference $= A_{15} - G_{15}$
$= 30.81404 - 30.68$
$= 0.13404$
$\approx 0.134\text{ km (shown)}$

1. Formulate the equation for the total distance covered in 13 weeks.

\text{Let the common ratio be } r = 1 + \frac{x}{100}.

\text{The weekly distances form a geometric progression with first term } a = 2.4 \text{ and number of terms } n = 13.

S_{13} = \frac{a(r^{13} - 1)}{r - 1}

170 = \frac{2.4(r^{13} - 1)}{r - 1}

2. Solve the equation for the common ratio $r$ using a graphing calculator.

\text{Using a graphing calculator to solve the equation }

\frac{2.4(r^{13} - 1)}{r - 1} = 170

r \approx 1.25426

3. Calculate the value of $x$.

1 + \frac{x}{100} = 1.25426

\frac{x}{100} = 0.25426

x = 25.426 \approx 25.4 \text{ (to 3 s.f.)}