IB Mathematics SL 1.8 The sum of infinite geometric sequences AA SL Paper 2- Exam Style Questions- New Syllabus
Line segment [AB] has length 2 cm and is divided into an infinite number of segments with lengths \( p \) cm, \( p^2 \) cm, \( p^3 \) cm, …, where \( 0 < p < 1 \).
Part (a):
Show that \( p = \frac{2}{3} \). [5]
Line segment [CD] has length \( b \) cm, where \( b > 1 \). Squares with side lengths \( k \) cm, \( k^2 \) cm, \( k^3 \) cm, …, where \( 0 < k < 1 \), are drawn along [CD] indefinitely.
Part (b):
Given that the total sum of the areas of all squares is \( \frac{9}{16} \), find \( b \). [9]
▶️ Answer/Explanation
Part (a)
Infinite sum of segments equals 2: \( p + p^2 + p^3 + \cdots = 2 \) (A1)
Recognize geometric series with first term \( p \), common ratio \( p \) (M1)
Use sum formula: \( S_\infty = \frac{p}{1 – p} \) (A1)
Set up equation: \( \frac{p}{1 – p} = 2 \) (A1)
Solve: \( p = 2 – 2p \implies 3p = 2 \implies p = \frac{2}{3} \) (A1) (AG)
[5 marks]
Part (b)
Recognize infinite geometric series for square areas: \( k^2 + k^4 + k^6 + \cdots \) (M1)
Use sum formula: \( S_\infty = \frac{k^2}{1 – k^2} = \frac{9}{16} \) (A2)
Solve: \( 16k^2 = 9 – 9k^2 \implies 25k^2 = 9 \implies k^2 = \frac{9}{25} \implies k = \frac{3}{5} \) (A1, A1)
Recognize segment lengths along [CD]: \( k + k^2 + k^3 + \cdots = b \) (M1)
Use sum formula: \( b = \frac{k}{1 – k} \) (A1)
Substitute \( k = \frac{3}{5} \): \( b = \frac{\frac{3}{5}}{1 – \frac{3}{5}} = \frac{\frac{3}{5}}{\frac{2}{5}} = \frac{3}{2} \) (M1, A1) (N3)
Answer: \( b = \frac{3}{2} \) [9 marks]