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IBDP Maths AA HL Topic - SL 1.3 Geometric sequences and series HL Paper 1- Exam Style Questions

IB DP Maths AA : HL Paper -1 : All Topics

Question

A collection of ten rectangular picture frames is labelled \( F_1, F_2, \ldots, F_9, F_{10} \).
The first frame \( F_1 \) has a width of \( 4 \,\text{cm} \) and a height of \( 5 \,\text{cm} \).
For each positive integer \( n \) with \( 1 \le n \le 9 \), both the width and the height of frame \( F_n \) are increased by \( 50\% \) to obtain the dimensions of the next frame \( F_{n+1} \).
(a) (i) Show that the area of frame \( F_n \) is \( 20 \left( \frac{9}{4} \right)^{n-1} \,\text{cm}^2 \).
(ii) Hence, determine the mean area of the ten picture frames, expressing your answer in the form \( p\!\left( \left( \frac{9}{4} \right)^a – 1 \right)\,\text{cm}^2 \), where \( p \in \mathbb{Q}^+ \) and \( a \in \mathbb{Z}^+ \).
(b) Determine 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 (IB Mathematics Analysis and Approaches 2021):

• SL 1.3: Geometric sequences and series — parts (a), (b)
• SL 4.3: Measures of central tendency (mean and median) — part (a)(ii), (b)
▶️ Answer/Explanation

(a)(i) Deriving the area formula:

for a sequence of areas, uses two consecutive terms to find a common
ratio OR for sequences of both widths and heights uses two consecutive
terms for both sequences to find both common ratios OR recognises that
both widths and heights are geometric sequences with commomn ratio.

Width of \( F_1 = 4 \) cm, height = 5 cm.
Increasing by 50% means multiplying by \( 1.5 = \frac{3}{2} \).
Thus:
Width of \( F_n = 4 \left( \frac{3}{2} \right)^{n-1} \)
Height of \( F_n = 5 \left( \frac{3}{2} \right)^{n-1} \)
Area = width × height = \( 4 \left( \frac{3}{2} \right)^{n-1} \times 5 \left( \frac{3}{2} \right)^{n-1} \)
= \( 20 \left( \frac{3}{2} \times \frac{3}{2} \right)^{n-1} = 20 \left( \frac{9}{4} \right)^{n-1} \) cm\(^2\).

(a)(ii) Mean area:

Areas form a geometric sequence with \( u_1 = 20 \), \( r = \frac{9}{4} \), \( n = 10 \).
Sum of areas: \( S_{10} = \frac{u_1(r^{10} – 1)}{r – 1} = \frac{20\left( \left( \frac{9}{4} \right)^{10} – 1 \right)}{\frac{9}{4} – 1} \)
\( \frac{9}{4} – 1 = \frac{5}{4} \), so denominator = \( \frac{5}{4} \).
Thus \( S_{10} = \frac{20\left( \left( \frac{9}{4} \right)^{10} – 1 \right)}{\frac{5}{4}} = 20 \times \frac{4}{5} \left( \left( \frac{9}{4} \right)^{10} – 1 \right) = 16 \left( \left( \frac{9}{4} \right)^{10} – 1 \right) \).
Mean = \( \frac{S_{10}}{10} = \frac{16}{10} \left( \left( \frac{9}{4} \right)^{10} – 1 \right) = \frac{8}{5} \left( \left( \frac{9}{4} \right)^{10} – 1 \right) \) cm\(^2\).
\( \boxed{p = \frac{8}{5}, \; a = 10} \)

(b) Median area:

With ten frames, the median is the average of the 5th and 6th areas.
Area of \( F_5 = 20 \left( \frac{9}{4} \right)^{4} \)
Area of \( F_6 = 20 \left( \frac{9}{4} \right)^{5} \)
Median = \( \frac{20 \left( \frac{9}{4} \right)^{4} + 20 \left( \frac{9}{4} \right)^{5}}{2} = 10 \left( \frac{9}{4} \right)^{4} \left( 1 + \frac{9}{4} \right) \)
\( 1 + \frac{9}{4} = \frac{13}{4} \)
So median = \( 10 \left( \frac{9}{4} \right)^{4} \times \frac{13}{4} = \frac{130}{4} \left( \frac{9}{4} \right)^{4} = \frac{65}{2} \left( \frac{9}{4} \right)^{4} \) cm\(^2\).
\( \boxed{q = \frac{65}{2}} \)

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