Home / IBDP Maths SL 1.2 Arithmetic Sequences & Series AA HL Paper 2- Exam Style Questions

IBDP Maths SL 1.2 Arithmetic Sequences & Series AA HL Paper 2- Exam Style Questions- New Syllabus

IB DP Maths AA HL Papers 2 - All Topics

Question

A pyramid structure is constructed using rectangular playing cards. The pyramid consists of \( n \) rows, where \( n \geq 1 \).
The construction involves placing some cards horizontally to act as supports, while others are tilted at an angle of \( 60^\circ \) to the horizontal.
The diagrams below illustrate the configuration of these pyramid stacks for \( n = 1 \), \( n = 2 \), and \( n = 3 \).
\( n = 1 \qquad \qquad n = 2 \qquad \qquad n = 3 \)
 
Let \( t_n \) denote the total number of cards required to build a pyramid with \( n \) rows.
(a) State the value of \( t_3 \).
(b) Find the value of \( t_4 \).
(c) Show that the general term for the number of cards is \( t_n = \frac{n(3n+1)}{2} \).
A standard deck contains 52 playing cards.
(d) If a complete pyramid is built using cards from exactly 14 full decks, determine the maximum possible number of rows in the stack.
(e) A complete pyramid is built using a minimum number of full decks such that every card from those decks is used and none remain. Calculate the minimum number of rows required for this to occur.
The length of the long edge of each card is 88 mm, as shown below:
 
(f) Determine the minimum number of cards required to build a complete pyramid stack with a vertical height exceeding 2 metres. (Ignore the thickness of the cards).

Syllabus Topic Codes (IB Mathematics AA HL):

• SL 1.2: Arithmetic sequences and series; use of the formulae for the \( n \)th term and the sum — parts (a), (b), (c) 
• SL 3.2: Use of sine, cosine and tangent ratios to find sides and angles of right-angled triangles — part (f) 
• AHL 2.12: Real and complex roots of polynomial equations — part (d)
▶️ Answer/Explanation

(a) From the diagram for \( n = 3 \), counting the cards gives: \( t_3 = 15 \).
\(\boxed{15}\)

(b) The increase in cards for each new row follows an arithmetic pattern.
Increase from \( n=1 \) to \( n=2 \): \( 7 – 2 = 5 \).
Increase from \( n=2 \) to \( n=3 \): \( 15 – 7 = 8 \).
Following the pattern, the next increase will be \( 8 + 3 = 11 \).
So, \( t_4 = 15 + 11 = 26 \).
\(\boxed{26}\)

(c) The number of cards in row \( k \) (counting from the top) is given by \( 2k \) (angled cards) + \( (k-1) \) (horizontal cards) = \( 3k – 1 \).
Total cards \( t_n = \sum_{k=1}^n (3k – 1) = 3 \left[ \frac{n(n+1)}{2} \right] – n \).
\( t_n = \frac{3n^2 + 3n – 2n}{2} = \frac{3n^2 + n}{2} = \frac{n(3n+1)}{2} \).
Shown.

(d) Total cards available: \( 14 \times 52 = 728 \).
Set \( \frac{n(3n+1)}{2} \leq 728 \implies 3n^2 + n – 1456 \leq 0 \).
Using the quadratic formula: \( n = \frac{-1 + \sqrt{1 + 4(3)(1456)}}{6} \approx 21.86 \).
Max rows: \(\boxed{21}\).

(e) We need \( t_n \) to be a multiple of 52: \( \frac{n(3n+1)}{2} = 52k \implies n(3n+1) = 104k \).
Testing integer values for \( n \):
If \( n = 13 \), \( t_{13} = \frac{13(39+1)}{2} = \frac{13 \times 40}{2} = 260 \).
Since \( 260 = 5 \times 52 \), no cards are left over.
Min rows: \(\boxed{13}\).

(f)
Vertical height of one row = \( 88 \sin 60^\circ = 88 \left( \frac{\sqrt{3}}{2} \right) = 44\sqrt{3} \) mm.
Total height \( H = n \times 44\sqrt{3} \).
Set \( n \times 44\sqrt{3} > 2000 \implies n > \frac{2000}{44\sqrt{3}} \approx 26.24 \).
Minimum rows \( n = 27 \).
Cards needed \( t_{27} = \frac{27(3 \cdot 27 + 1)}{2} = \frac{27 \times 82}{2} = 1107 \).
\(\boxed{1107}\)

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