Dendrologists are specialised scientists who study trees and woody plants. Their work is diverse and can encompass various activities related to the identification, classification, biology, and ecology of trees. A group of dendrologists are studying the growth of 2 species of trees, codenamed Tree Vee and Tree Jay. In the $1^{\text{st}}$ year, the height of Tree Vee and Tree Jay are both $H$ cm.

1. Find the increase in height for the 3rd and 4th years.

The increases form a geometric progression with a common ratio of $1 - 0.10 = 0.9$.
Increase in $3^{\text{rd}}$ year $= 0.9s$
Increase in $4^{\text{th}}$ year $= 0.9(0.9s) = 0.81s$

2. Calculate the total height in the 4th year by adding the initial height and the yearly increases.

$\text{Height in } 4^{\text{th}} \text{ year} = H + s + 0.9s + 0.81s$
$= H + (1 + 0.9 + 0.81)s$
$= H + 2.71s \text{ (shown)}$

1. Identify the sequence of annual height increases as a geometric progression.

The increases in height each year form a geometric progression with the first term $a = s$ and a common ratio $r = 1 - 0.10 = 0.9$.

2. Relate the tree's height in the $n^{\text{th}}$ year to the sum of the increases.

By the $n^{\text{th}}$ year, there have been $(n-1)$ years of height increases. The total height is the initial height $H$ plus the sum of these $(n-1)$ terms.
$\text{Height in } n^{\text{th}} \text{ year} = H + S_{n-1}$

3. Apply the geometric series sum formula and simplify to the given expression.

Using the formula for the sum of a geometric progression, $S_k = \frac{a(1 - r^k)}{1 - r}$:
$S_{n-1} = \frac{s(1 - 0.9^{n-1})}{1 - 0.9}$
$S_{n-1} = \frac{s(1 - 0.9^{n-1})}{0.1} = 10s(1 - 0.9^{n-1})$
Therefore, the total height is:
$H + 10s(1 - 0.9^{n-1}) \text{ (shown)}$

1. Evaluate the expression for height as $n$ approaches infinity to find the maximum height

As $n \to \infty$, $0.9^{n-1} \to 0$.
Therefore, the maximum height is $H + 10s(1 - 0) = H + 10s$.

1. Identify the arithmetic progression for the annual height increases.

The annual increases in height from the $2^{\text{nd}}$ year onwards form an arithmetic progression.
First term, $a = t$
Common difference, $d = -0.5$
Number of increases from the $2^{\text{nd}}$ to the $10^{\text{th}}$ year, $n = 10 - 1 = 9$

2. Calculate the total increase in height using the sum formula for an arithmetic progression.

$\begin{aligned} S_9 &= \frac{9}{2} \left[2a + (9-1)d\right] \\ &= 4.5 \left[2(t) + 8(-0.5)\right] \\ &= 4.5 (2t - 4) \\ &= 9t - 18 \end{aligned}$

3. Find the total height in the 10th year by adding the initial height to the total increase.

$\begin{aligned} \text{Height in } 10^{\text{th}} \text{ year} &= \text{Height in } 1^{\text{st}} \text{ year} + S_9 \\ &= H + 9t - 18 \\ &= (H - 18 + 9t) \text{ cm (shown)} \end{aligned}$

1. Calculate the height of Tree *Jay* in the $10^{\text{th}}$ year by substituting $t = 20$ into the expression from part (d).

$\text{Height in } 10^{\text{th}} \text{ year} = H - 18 + 9(20)$
$= H - 18 + 180$
$= H + 162$

2. Formulate the height for the $n^{\text{th}}$ year by adding the constant growth over the additional $(n - 10)$ years, and simplify.

$\text{Height in } n^{\text{th}} \text{ year} = (H + 162) + 7(n - 10)$
$= H + 162 + 7n - 70$
$= H + 7n + 92$

1. Find the expression for the height of Tree Vee in the $n$th year.

From (b), Tree Vee's height in the $n^\text{th}$ year is $V_n = H + 10s(1 - 0.9^{n-1})$.

Substitute $s = 30$:
$V_n = H + 10(30)(1 - 0.9^{n-1}) = H + 300(1 - 0.9^{n-1})$

2. Derive the expression for the height of Tree Jay in the $n$th year for $n \ge 11$.

From (d), Tree Jay's height in the $10^\text{th}$ year is $H - 18 + 9(20) = H + 162$.

Since Tree Jay's height increases by a constant 7 cm per year after the $10^\text{th}$ year, its height in the $n^\text{th}$ year ($n \ge 11$) is given by an arithmetic progression:
$J_n = H + 162 + 7(n - 10)$
$J_n = H + 7n + 92$

3. Formulate the inequality for the difference in heights and solve for $n$ using a graphing calculator.

The trees' heights are within 7 cm of each other, so $|V_n - J_n| \le 7$.

$|H + 300(1 - 0.9^{n-1}) - (H + 7n + 92)| \le 7$
$|300 - 300(0.9^{n-1}) - 7n - 92| \le 7$
$|208 - 7n - 300(0.9^{n-1})| \le 7$

Let $f(n) = 208 - 7n - 300(0.9^{n-1})$. Using a graphing calculator's table function to check integer values of $n \ge 12$ (since the year is after 2034):
When $n = 25$, $f(25) \approx 9.08$
When $n = 26$, $f(26) \approx 4.47 \implies |f(26)| \le 7$
When $n = 27$, $f(27) \approx -0.38 \implies |f(27)| \le 7$
When $n = 28$, $f(28) \approx -5.44 \implies |f(28)| \le 7$
When $n = 29$, $f(29) \approx -10.69$

Thus, the condition is satisfied for $n = 26$, $n = 27$, and $n = 28$.

4. Convert the values of $n$ to calendar years.

The $1^\text{st}$ year corresponds to 2024. Therefore, the $n^\text{th}$ year corresponds to the calendar year $(2024 + n - 1) = (2023 + n)$.

For $n = 26$, the year is $2023 + 26 = 2049$.
For $n = 27$, the year is $2023 + 27 = 2050$.
For $n = 28$, the year is $2023 + 28 = 2051$.

These years are all after 2034.