Out-and-back circuits and changing growth rates · source reference

Printed question

Besto’s coach designs a training programme starting with a 400m run on session 1. Each subsequent session covers 250m more than the previous.

(i) Find the minimum sessions for at least 20km in one session. [3]

For another group in Besto, a stamina circuit starts at O with points $P_1,P_2,P_3,\ldots$ increasingly far away on a straight line. Stage 1 goes O to $P_1$ and back, stage 2 O to $P_2$ and back, etc. The diagram (Fig. 1) gives $OP_1=50$m, $P_1P_2=100$m, $P_2P_3=300$m and $P_nP_{n+1}=3P_{n-1}P_n$.

(ii) Find distance after $n$ stages. [3]

(iii) Hence find distance from O and travel direction after exactly 42km. [3]

Choco’s coach designs another programme starting at 400m in session 1. Each subsequent distance increases by 10% of the previous session’s distance. From session 11 onward and for every subsequent session, distance increases by $r$% of the previous distance.

(iv) If Choco runners cover at least 20km in session 70, find the range of $r$. [4]

Printed working

(i) Let $T_n$ denote distance covered in session $n$. It is an AP with difference 250:

$$T_n=400+250(n-1)\ge20000.$$

GC table below gives minimum $n=80$.

Alternative: $250(n-1)\ge19600\Rightarrow n-1\ge78.4\Rightarrow n\ge79.4$. Minimum 80.

(ii) From the stage table below,

$$\begin{aligned}\text{total after }n\text{ stages}&=2(50)[1+3+3^2+\cdots+3^{n-1}]\\&=100\frac{1(3^n-1)}{3-1}=50(3^n-1).\end{aligned}$$

(iii) Completed stages satisfy $50(3^n-1)\le42000$. Table below shows six completed stages, 36400m. Remaining $42000-36400=5600$m. Since $OP_7=50\times3^6=36450>5600$, runner is moving away from O, 5600m from O, and has not reached $P_7$.

(iv) Session 10: $400(1.1)^9$m. From session 11 on, session $n$ distance $400(1.1)^9(1+r/100)^{n-10}$.

$$400(1.1)^9(1+r/100)^{60}\ge20000\Rightarrow(1+r/100)^{60}\ge21.205.$$

Let $x=r/100$. GC intersection of $y=(1+x)^{60}$ and $y=21.205$ is $x\approx0.0522218$. Thus $r/100\ge0.05222\Rightarrow r\ge5.222=5.22$ (3 s.f.).

Source tables and calculator displays

$n$	$T_n$
79	19900 <20000
80	20150 >20000
81	20400 >20000

$n$	Total stage distance expansion
1	$2(50)$
2	$2(50)+2(150)=2(50)+2(3)(50)=2(50)[1+3]$
3	$2(50)+2(150)+2(450)=2(50)+2(3)(50)+2(3)^2(50)=2(50)[1+3+3^2]$
…	…
$n$	$2(50)+2(3)(50)+\cdots+2(3)^{n-1}(50)$

$n$	$50(3^n-1)$
5	12100 <42000
6	36400 <42000
7	109300 >42000

Source diagrams redrawn

Printed answer key

(i) 80; (ii) $50(3^n-1)$; (iii) running away from O at distance 5600m; (iv) $r\ge5.22$ (3 s.f.)

Mathematical review and qualifications

(i) $400+250(n-1)\ge20000$ gives $n\ge79.4$, so first session 80.

(ii) $OP_j=50+100\sum_{k=0}^{j-2}3^k=50\cdot3^{j-1}$ (including $j=1$). Total out-and-back distance is $2\sum_{j=1}^nOP_j=50(3^n-1)$ metres.

(iii) Six stages cover $50(3^6-1)=36400$ m, while seven cover 109300 m. Remaining travel is $42000-36400=5600$ m in stage 7. Since $OP_7=36450$ m, the runner is still on the outward leg, 5600 m from O and running away.

(iv) Session 10 distance is $400(1.1)^9$. Session 70 distance is $400(1.1)^9(1+r/100)^{60}\ge20000$. For the intended increase $r\ge0$, this gives $r\ge100[(50/1.1^9)^{1/60}-1]\approx5.22217\%$, reported as $r\ge5.22\%$ (3 s.f.) with the exact lower bound retained.

Nine increases reach session 10; sixty further increases reach session 70. Rounded lower bounds are approximate and should not replace the exact inequality in calculations.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2022/SAJC/Prelim/I/9] Question PDF pages 7; solution PDF pages 30, 31, 32. The original files remain private QA records outside this native content.