Exercise 8.3 Practice

We are given common ratio $r = 2$ and $t_8 = 192$.
The $n$-th term of a GP is $t_n = a \cdot r^{n-1}$.
$$t_8 = a \cdot r^7 \implies a \cdot 2^7 = 192$$
Since $2^7 = 128$:
$$a \cdot 128 = 192 \implies a = \frac{192}{128} = \mathbf{1.5}$$

Now calculate the 12th term ($t_{12}$):
$$t_{12} = a \cdot r^{11} = 1.5 \times 2^{11}$$
Since $2^{11} = 2048$:
$$t_{12} = 1.5 \times 2048 = \mathbf{3072}$$

Identify first term $a = 5$ and common ratio $r = \frac{25}{5} = 5$.

$n$-th Term ($t_n$):
$$t_n = a \cdot r^{n-1} = 5 \cdot 5^{n-1} = \mathbf{5^n}$$

10th Term ($t_{10}$):
$$t_{10} = 5^{10} = \mathbf{9765625}$$

Let's compute the first few terms of the sequence to identify a pattern:
• $t_1 = 2$
• $t_2 = 3(2) - 2 = 4$
• $t_3 = 3(4) - 2 = 10$
• $t_4 = 3(10) - 2 = 28$
• $t_5 = 3(28) - 2 = 82$

Notice the relationship to powers of 3:
• $t_1 = 3^0 + 1 = 2$
• $t_2 = 3^1 + 1 = 4$
• $t_3 = 3^2 + 1 = 10$
• $t_4 = 3^3 + 1 = 28$
• $t_5 = 3^4 + 1 = 82$
General term: $t_n = 3^{n-1} + 1$.

Set $t_n = 730$ and solve for $n$:
$$3^{n-1} + 1 = 730$$
$$3^{n-1} = 729$$
Since $729 = 3^6$:
$$n - 1 = 6 \implies n = 7$$

Identify first term $a = 2$ and common ratio $r = 3$.
The explicit formula is:
$$t_n = 2 \cdot 3^{n-1}$$
The recursive formula is:
$$t_1 = 2, \quad t_n = 3t_{n-1} \text{ for } n \ge 2$$

Set $t_n = 4374$ and solve for $n$:
$$2 \cdot 3^{n-1} = 4374$$
$$3^{n-1} = 2187$$
Since $2187 = 3^7$:
$$n - 1 = 7 \implies n = 8$$

(i) Height after 5th bounce:
Initial height $H_0 = 80\text{ m}$. Common ratio $r = 0.60$.
$$H_5 = H_0 \cdot r^5 = 80 \times (0.6)^5$$
$$(0.6)^5 = 0.07776$$
$$H_5 = 80 \times 0.07776 = \mathbf{6.22\text{ metres}}$$

(ii) Total Vertical Distance by 6th Ground Hit:
The path consists of the initial drop ($80\text{ m}$), followed by 5 bounces. Each bounce has a rise and a fall of equal height:
$$\text{Distance} = H_0 + 2(H_1 + H_2 + H_3 + H_4 + H_5)$$
Where $H_k = 80 \times (0.6)^k$.
Let's calculate each height:
• $H_1 = 80 \times 0.6 = 48\text{ m}$
• $H_2 = 48 \times 0.6 = 28.8\text{ m}$
• $H_3 = 28.8 \times 0.6 = 17.28\text{ m}$
• $H_4 = 17.28 \times 0.6 = 10.368\text{ m}$
• $H_5 = 10.368 \times 0.6 = 6.2208\text{ m}$

Sum of these heights:
$$S_{heights} = 48 + 28.8 + 17.28 + 10.368 + 6.2208 = 110.6688\text{ m}$$
Total distance:
$$\text{Total Distance} = 80 + 2(110.6688) = 80 + 221.3376 = \mathbf{301.34\text{ metres}}$$

The sequence is $t_n = 2^{n+1}$ for $n \ge 1$:
• $t_1 = 2^2 = 4$
• $t_2 = 2^3 = 8$
• $t_3 = 2^4 = 16$

Set $t_n = 128$:
$$2^{n+1} = 128$$
Since $128 = 2^7$:
$$n + 1 = 7 \implies n = 6$$

(i) Red Squares count for Stages 0 to 3:
• Stage 0: **$1$**
• Stage 1: **$8$** (since 1 is removed out of 9)
• Stage 2: $8 \times 8 = \mathbf{64}$
• Stage 3: $64 \times 8 = \mathbf{512}$

(ii) Predictions:
• Stage 4: $512 \times 8 = \mathbf{4096}$
• Stage 5: $4096 \times 8 = \mathbf{32768}$

(iii) Rules for number of red squares ($N_n$):
• **Explicit:** $N_n = 8^n$ for $n \ge 0$.
• **Recursive:** $N_0 = 1, \quad N_n = 8N_{n-1}$ for $n \ge 1$.

(iv) Red Area ($A_n$):
At each stage, the area is multiplied by $\frac{8}{9}$ (since 1/9 of the remaining area is removed).
• Stage 1 Area: $\mathbf{\frac{8}{9} \approx 0.889}$
• Stage 2 Area: $\mathbf{\left(\frac{8}{9}\right)^2 = \frac{64}{81} \approx 0.790}$
• Stage 3 Area: $\mathbf{\left(\frac{8}{9}\right)^3 = \frac{512}{729} \approx 0.702}$
• Stage 4 Area: $\mathbf{\left(\frac{8}{9}\right)^4 = \frac{4096}{6561} \approx 0.624}$
• Stage 5 Area: $\mathbf{\left(\frac{8}{9}\right)^5 = \frac{32768}{59049} \approx 0.555}$
• **Explicit Area Rule:** $A_n = \left(\frac{8}{9}\right)^n$
• **Recursive Area Rule:** $A_0 = 1, \quad A_n = \frac{8}{9} A_{n-1}$
• **As $n \to \infty$:** Since $\frac{8}{9} < 1$, the area of the red region converges to **$0$** as the number of stages increases indefinitely.