IB Mathematics AHL 1.11 The sum of infinite geometric sequences AI HL Paper 2- Exam Style Questions- New Syllabus

a
Method 1: Area of Sectors
Area of first sector: \( \frac{1}{2} \cdot 2^2 \cdot \theta = 2\theta \)
Areas: \( 2\theta, 2k\theta, 2k^2\theta, \ldots \)
Sum: \( 2\theta (1 + k + k^2 + \cdots) = 2\theta \cdot \frac{1}{1 – k} = 4\pi \)
\( \theta = 2\pi (1 – k) \)
Method 2: Angles of Sectors
Angles: \( \theta, k\theta, k^2\theta, \ldots \)
Sum: \( \theta (1 + k + k^2 + \cdots) = \theta \cdot \frac{1}{1 – k} = 2\pi \)
\( \theta = 2\pi (1 – k) \)
Result: \( \theta = 2\pi (1 – k) \) [5].

b
Perimeter of first sector: \( 4 + 2\theta \)
Perimeter of third sector: \( 4 + 2 k^2 \theta \)
Given: \( 4 + 2 k^2 \theta = \frac{1}{2} (4 + 2\theta) = 2 + \theta \)
Simplify: \( \theta = \frac{2}{1 – 2 k^2} \)
Using \( \theta = 2\pi (1 – k) \): \( 2\pi (1 – k) = \frac{2}{1 – 2 k^2} \)
\( \pi (1 – k)(1 – 2 k^2) = 1 \)
Solve cubic: \( 2 k^3 – 2 k^2 – k + 1 – \frac{1}{\pi} = 0 \)
\( k \approx 0.456 \), \( \theta = 2\pi (1 – 0.456) \approx 3.42 \)
Result: \( k = 0.456 \), \( \theta = 3.42 \) [7].