Selected terms and a rabbit–fox chase · source reference

Printed question

(a) An AP has first term $a>0$ and difference $d\ne0$. Its eighth, third and second terms are the first three terms of an infinite GP.

(i) Find its common ratio. [3]

(ii) Find the exact sum of its odd-numbered terms in terms of $a$. [3]

(b) A programmer codes a rabbit-fox chase along a straight path to model an actual hunt. The rabbit first hops 1.75m, then 1% less each hop. The fox first leaps 3m, then 0.02m less each leap. Initially the rabbit is 60m ahead. They start and end each hop/leap at the same time.

(i) Find the total distances after $n$ leaps/hops, and the minimum number for catch-up. [4]

(ii) Find fox leaps before stopping. Hence find the minimum starting distance in metres such that the fox never catches the rabbit, to the nearest integer. [2]

Printed working

(a)(i) Let $b,r$ be GP first term and common ratio:

$$b=a+7d\ (1),\quad br=a+2d\ (2),\quad br^2=a+d\ (3).$$

$(1)-(2)$ gives $b-br=5d$ (4); $(2)-(3)$ gives $br-br^2=d$ (5).

Divide (4) by (5):

$$\frac{1-r}{r-r^2}=5\Rightarrow5r^2-6r+1=0\Rightarrow(5r-1)(r-1)=0.$$

$r=1/5$ or 1 (reject since $d\ne0$).

(ii) From (5), $\frac4{25}b=d$; from (3), $b=-\frac{25}3a$.

$$S_{\text{odd}}=\frac b{1-r^2}=\frac{-25a/3}{1-1/25}=-\frac{625a}{72}.$$

(b)(i)

$$S_{\text{fox}}=\frac n2[2(3)+(n-1)(-0.02)]=n(3.01-0.01n),$$ $$S_{\text{rabbit}}=\frac{1.75(1-0.99^n)}{1-0.99}=175(1-0.99^n).$$

For catch-up $n(3.01-0.01n)-175(1-0.99^n)\ge60$. Let $Y$ be the left side; GC table below gives least $n=54$.

(ii) Let $k$ be the starting gap. No catch-up requires

$$k>\max[n(3.01-0.01n)-175(1-0.99^n)].$$

Fox stops when $T_n=0$:

$$3+(n-1)(-0.02)=0\Rightarrow n=151.$$

It takes 150 positive leaps. For $0\le n\le150$, the GC maximum is approximately 94.827. Hence minimum whole-metre gap 95m.

Teacher notes

Eliminate $a$ first. Then eliminate $b,d$ simultaneously by dividing (4) by (5); the final equation contains $r$ only.

Odd-indexed GP terms $T_1,T_3,T_5,\ldots=b,br^2,br^4,\ldots$, first $b$, ratio $r^2$.

Fox: $3,3-0.02,3-2(0.02),\ldots$ is AP ($a=3,d=-0.02$). Rabbit: $1.75,1.75(0.01),1.75(0.01)^2,\ldots$ and ratio $r=0.01$ as printed in the margin. This is a source typo: the question and working consistently use 0.99, since distance decreases by 1%.

Refer to Q15(iv) for GC table skills. The gap calculation is distance travelled by fox minus distance travelled by rabbit.

Source tables and calculator displays

$n$	$Y$
53	59.171 <60
54	60.084 >60
55	60.987 >60

Source diagrams redrawn

Printed answer key

(a)(i) $1/5$; (ii) $-625a/72$; (b)(i) 54; (ii) 150 leaps; 95m

Mathematical review and qualifications

(a)(i) $(a+2d)^2=(a+7d)(a+d)$ gives $d(4a+3d)=0$. As $d \ne0$, $d=-4a/3$ and $r=1/5$. (ii) The GP first term is $a+7d=-25a/3$. Its odd-indexed sum is $(-25a/3)/(1-1/25)=-625a/72$.

(b)(i) $F_n=n[6-0.02(n-1)]/2=n(3.01-0.01n)$ and $R_n=175(1-0.99^n)$. Catch-up requires $F_n-R_n\ge60$. The values at 53 and 54 are about 59.171 and 60.084; first count is 54.

(ii) The fox nth leap is $3-0.02(n-1)$, zero at $n=151$, hence 150 positive leaps. Checking integer $0\le n\le150$ gives maximum gain about $94.8256$ m at $n=126$. A gap strictly above this is safe in the endpoint model; the minimum whole-metre gap is 95 m. The source continuous-index maximum is approximately 94.827, consistent with the same integer answer.

The model specifies synchronised endpoints. Within-hop motion is unspecified; the distance calculation follows the source endpoint/constant-progress model. A strict real lower bound has an infimum rather than an attained minimum.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2022/ASRJC/Prelim/I/9] Question PDF pages 5; solution PDF pages 23, 24. The original files remain private QA records outside this native content.