Home / IB Mathematics SL 4.3 Measures of central tendency AA SL Paper 1- Exam Style Questions

IB Mathematics SL 4.3 Measures of central tendency AA SL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AA SL Paper 1 -All Topics

Question

A series of ten rectangular picture frames \( F_1, F_2, \dots, F_{10} \) is constructed. The first frame \( F_1 \) has width \(4\) cm and height \(5\) cm. For each frame \( F_n \), both its width and its height are increased by \(50\%\) to produce the next frame \( F_{n+1} \) ( \( n = 1,2,\dots,9\) ).
(a) (i) Show that the area of picture frame \( F_n \) is \( 20\left( \frac{9}{4} \right)^{\,n-1}\ \text{cm}^2 \).
(ii) Hence, find the mean area of the ten picture frames, giving your answer in the form \( p\Bigl( \bigl( \tfrac{9}{4} \bigr)^a – 1 \Bigr)\ \text{cm}^2 \), where \( p \in \mathbb{Q}^+ \), \( a \in \mathbb{Z}^+ \).
(b) Find the median area of the ten picture frames, giving your answer in the form \( q\left( \frac{9}{4} \right)^4\ \text{cm}^2 \), where \( q \in \mathbb{Q}^+ \).

Most-appropriate topic codes (Mathematics: analysis and approaches guide):

• SL 1.3: Geometric sequences and series; Use of the formulae for the \(n^{th}\) term and the sum of the first \(n\) terms — Part a(i), a(ii) 
• SL 4.3: Measures of central tendency (mean, median and mode) — Part a(ii), b 
▶️ Answer/Explanation

(a)(i) Area of \(F_n\):

Width of \(F_1 = 4\) cm, height = 5 cm.
Each dimension grows by 50% each step ⇒ common ratio \(r = 1.5 = \frac{3}{2}\).
Thus:
Width of \(F_n = 4 \times \left(\frac{3}{2}\right)^{n-1}\)
Height of \(F_n = 5 \times \left(\frac{3}{2}\right)^{n-1}\)
Area = width × height = \( \left[4\left(\frac{3}{2}\right)^{n-1}\right] \times \left[5\left(\frac{3}{2}\right)^{n-1}\right] = 20 \left(\frac{9}{4}\right)^{n-1}\ \text{cm}^2. \)

(a)(ii) Mean area:

Areas form a geometric sequence with first term \(u_1 = 20\), common ratio \(R = \frac{9}{4}\).
Sum of ten terms: \( S_{10} = \frac{20\left(\left(\frac{9}{4}\right)^{10} – 1\right)}{\frac{9}{4} – 1} = \frac{20\left(\left(\frac{9}{4}\right)^{10} – 1\right)}{\frac{5}{4}} = 20 \times \frac{4}{5} \left[\left(\frac{9}{4}\right)^{10} – 1\right] = 16\left[\left(\frac{9}{4}\right)^{10} – 1\right]. \)
Mean = \( \dfrac{S_{10}}{10} = \dfrac{16}{10}\left[\left(\frac{9}{4}\right)^{10} – 1\right] = \dfrac{8}{5}\left[\left(\frac{9}{4}\right)^{10} – 1\right]. \)
Hence \( p = \dfrac{8}{5},\ a = 10 \).

(b) Median area:

For ten ordered areas, median is the average of the 5th and 6th terms.
5th term: \( u_5 = 20\left(\frac{9}{4}\right)^{4} \)
6th term: \( u_6 = 20\left(\frac{9}{4}\right)^{5} \)
Median = \( \frac{u_5 + u_6}{2} = \frac{20\left(\frac{9}{4}\right)^{4} + 20\left(\frac{9}{4}\right)^{5}}{2} = 10\left(\frac{9}{4}\right)^{4}\left(1 + \frac{9}{4}\right) = 10\left(\frac{9}{4}\right)^{4} \times \frac{13}{4} = \frac{130}{4}\left(\frac{9}{4}\right)^{4} = \frac{65}{2}\left(\frac{9}{4}\right)^{4}. \)
Hence \( q = \dfrac{65}{2} \).

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