Arithmetic and geometric progressions: ACJC 2025 Promo Q11

(a)

The floors form an arithmetic progression with $a = 10$ and $d = 2$: $$\begin{aligned} u_{10} &= 10 + 9(2) = 28 \text{ floors}. \end{aligned}$$

(b)

$$\begin{aligned} 10 + (n-1)(2) &= 62\\ n &= 27. \end{aligned}$$ He trains $3$ days a week, so $k = \dfrac{27}{3} = 9$ weeks.

(c)

Over $20$ weeks he has $60$ training days. For the first $27$ days, $$\begin{aligned} S_{27} &= \frac{27}{2}(10 + 62) = 972. \end{aligned}$$ For the remaining $33$ days he completes $62$ floors each: $$\begin{aligned} \text{Total} &= 972 + 62(33) = 3018 \text{ floors}. \end{aligned}$$

(d)

The heart rate forms a geometric progression with $a = 180$ and $r = 0.9$: $$\begin{aligned} 180(0.9)^{n-1} &< 82\\ (0.9)^{n-1} &< \frac{41}{90}\\ (n-1)\ln 0.9 &< \ln\frac{41}{90}\\ n - 1 &> 7.46. \end{aligned}$$ Hence $n > 8.46$, so the least value is $n = 9$.

Dividing by $\ln 0.9$ reverses the inequality, since $\ln 0.9$ is negative.

(e)

Using $H_{n+1} = 0.85H_n + 10.5$ with $H_1 = 170$: $H_{10} = 93.2$ and $H_{11} = 89.7$. The least value is $n = 11$.

(f)

In the long run the heart rate stabilises, so $H = 0.85H + 10.5$: $$\begin{aligned} 0.15H &= 10.5\\ H &= 70 \text{ bpm}. \end{aligned}$$

The solution uses the question’s initial heart rate of 170 bpm; the printed solution’s initial value of 160 was a typo.