IB Mathematics SL 1.2 Arithmetic Sequences & Series AA SL Paper 2- Exam Style Questions- New Syllabus

(a)
From the diagram (or by counting), a 3‑row pyramid uses \( t_3 = 15 \) cards. \( \boxed{15} \).

(b)
The sequence \( t_n \) increases by adding a new row of cards. Observing the pattern:
\( t_1 = 2 \), \( t_2 = 7 \), \( t_3 = 15 \).
The difference \( t_4 – t_3 \) is the number of cards in the 4th row, which is \( 3 \times 4 – 1 = 11 \) (or from the arithmetic pattern).
Thus \( t_4 = 15 + 11 = 26 \). \( \boxed{26} \).

(c)
Method 1 (Arithmetic series)
The number of cards in row \( r \) is \( 3r – 1 \).
Thus \( t_n = \sum_{r=1}^{n} (3r – 1) = 3\sum_{r=1}^{n} r – \sum_{r=1}^{n} 1 = 3 \cdot \frac{n(n+1)}{2} – n \).
Simplify: \( t_n = \frac{3n^2 + 3n – 2n}{2} = \frac{3n^2 + n}{2} = \frac{n(3n+1)}{2} \). \( \boxed{\frac{n(3n+1)}{2}} \).

(d)
Total cards available: \( 14 \times 52 = 728 \).
We need \( t_n \leq 728 \).
Solve \( \frac{n(3n+1)}{2} \leq 728 \Rightarrow 3n^2 + n – 1456 \leq 0 \).
Discriminant: \( \Delta = 1 + 4 \cdot 3 \cdot 1456 = 17473 \).
Positive root: \( n = \frac{-1 + \sqrt{17473}}{6} \approx \frac{-1 + 132.18}{6} \approx 21.86 \).
Thus the largest integer \( n \) satisfying the inequality is \( n = 21 \).
Check: \( t_{21} = \frac{21 \cdot 64}{2} = 672 \), \( t_{22} = \frac{22 \cdot 67}{2} = 737 \). \( \boxed{21} \).

(e)
We need \( t_n \) to be a multiple of 52.
Set \( \frac{n(3n+1)}{2} = 52k \) for some positive integer \( k \).
Trial of small \( n \):
\( n = 5 \): \( t_5 = 40 \)
\( n = 8 \): \( t_8 = 100 \)
\( n = 13 \): \( t_{13} = \frac{13 \cdot 40}{2} = 260 = 5 \times 52 \).
Thus the smallest \( n \) giving a multiple of 52 is \( n = 13 \). \( \boxed{13} \).

(f)
In the pyramid, each slanted card forms the side of an equilateral triangle of side 88 mm. The vertical height of such a triangle is \( 88 \sin 60^\circ = 88 \cdot \frac{\sqrt{3}}{2} = 44\sqrt{3} \) mm.
With \( n \) rows, the total vertical height is \( 44\sqrt{3} \times n \) mm.
We require \( 44\sqrt{3} n > 2000 \) mm.
Solve: \( n > \frac{2000}{44\sqrt{3}} \approx 26.243 \).
Thus the smallest integer \( n \) is \( n = 27 \).
Now find the number of cards for \( n = 27 \):
\( t_{27} = \frac{27 \cdot (81+1)}{2} = 1107 \). \( \boxed{1107} \).