Adam and Gregory signed up for a marathon. In preparation for this marathon, Adam and Gregory each planned a 15-week personalised training programme. Adam runs 2.4 km on the first day of Week 1, and on the first day of each subsequent week, the distance covered is increased by 20% of the previous week. Gregory also runs 2.4 km on the first day of Week 1, but on the first day of each subsequent week, the distance covered is increased by $d$ km, where $d$ is a constant. Assume Adam and Gregory only run on the first day of each week.

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.)}