Porisma

Worked solution

Compare AP and GP travel totals in a chase

A zoology student models a jaguar chasing a white-tailed deer initially dd metres ahead. Both start together. The jaguar leaps 66 m first, each later leap 0.10.1 m shorter. The deer leaps 99 m first, each later leap 5%5\% shorter.

(i) Find the deer’s total after nn leaps and deduce its maximum distance.

(ii) Assume both complete the same number of leaps in the same duration. For d=11d=11 m, find the least nn for the jaguar to catch the deer within nn leaps.

[2018/MJC/I/10(b)]

CJC · Sequences and Series · Example 34 (Self-Reading)

(i) Deer distance and limiting bound

Answer

Dn=180(1−0.95n)D_n=180(1-0.95^n) metres; the limiting upper bound is 180180 m.

Sum the shrinking geometric leap lengths

Dn=9(1−0.95n)1−0.95=180(1−0.95n)\begin{aligned}&D_n\\ & =\frac{9(1-0.95^n)}{1-0.95}\\ & =180(1-0.95^n)\end{aligned}

First leap 9 and ratio 0.95 give the finite sum.

Interpret the word maximum

0<0.95n⟹Dn<180,lim⁡n→∞Dn=1800<0.95^n\quad\Longrightarrow\quad D_n<180,\qquad\lim_{n\to\infty}D_n=180

As n grows, 0.95^n tends to zero and D_n approaches 180. The source calls this the maximum distance; it is a supremum, not a distance attained after a finite number of leaps.

How did this part go?

(ii) First catch with an eleven-metre gap

Answer

n=50n=50.

Compare total distances at the same leap count

Jn=n2[12+(n−1)(−0.1)],Jn≥Dn+11J_n=\frac n2[12+(n-1)(-0.1)],\qquad J_n\ge D_n+11

The jaguar AP has first term 6 and difference −0.1. It must cover the deer’s travelled distance plus the initial 11 m gap.

Use the calculator difference table

F(n)=n2[12+(n−1)(−0.1)]−180(1−0.95n)−11,F(49)≈−0.021<0,F(50)≈0.350>0,F(51)≈0.658>0.\begin{aligned}F(n)&=\frac n2[12+(n-1)(-0.1)]-180(1-0.95^n)-11,\\F(49)&\approx-0.021<0,\\F(50)&\approx0.350>0,\\F(51)&\approx0.658>0.\end{aligned}

Enter the source function and inspect positive integer n. The source written table shows the adjacent crossing values.

Select the first feasible integer

F(n)<0 (1≤n≤49),F(50)>0⟹nmin⁡=50F(n)<0\ (1\le n\le49),\qquad F(50)>0\quad\Longrightarrow\quad n_{\min}=50

Checking earlier integer counts 1–49 gives negative differences; 50 is the first nonnegative one. This count is within the model’s positive jaguar-leap range, 1–60.

How did this part go?

Common mistakes

Compare cumulative distances and include the initial separation. Do not compare only the nth leap lengths or extend an AP into negative travel distances.