IBDP Maths SL 1.2 Arithmetic Sequences & Series AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) The \(k\)-th term of an arithmetic sequence is given by \( u_k = u_1 + (k-1)d \), where \( u_1 = 60 \) and \( d = -2.5 \).
Set \( u_k = 0 \): \( 60 + (k-1)(-2.5) = 0 \) M1.
Solve: \( 60 – 2.5(k-1) = 0 \Rightarrow 60 = 2.5(k-1) \Rightarrow k-1 = \frac{60}{2.5} = 24 \Rightarrow k = 25 \) A1.
[2 marks]
(b) The sum of the first \( n \) terms of an arithmetic sequence is \( S_n = \frac{n}{2} [2u_1 + (n-1)d] \).
Substitute \( u_1 = 60 \), \( d = -2.5 \): \( S_n = \frac{n}{2} [2 \cdot 60 + (n-1)(-2.5)] = \frac{n}{2} [120 – 2.5(n-1)] \) M1.
Simplify: \( S_n = \frac{n}{2} [122.5 – 2.5n] = n [61.25 – 1.25n] \).
To find the maximum, consider \( S_n \) as a quadratic function in \( n \): \( S_n = -1.25n^2 + 61.25n \). The vertex of a quadratic \( an^2 + bn \) occurs at \( n = -\frac{b}{2a} \).
Here, \( a = -1.25 \), \( b = 61.25 \): \( n = -\frac{61.25}{2 \cdot (-1.25)} = \frac{61.25}{2.5} = 24.5 \).
Since \( n \) is an integer, evaluate at \( n = 24 \) and \( n = 25 \):
For \( n = 24 \): \( S_{24} = \frac{24}{2} [120 – 2.5 \cdot 23] = 12 \cdot [120 – 57.5] = 12 \cdot 62.5 = 750 \).
For \( n = 25 \): \( S_{25} = \frac{25}{2} [120 – 2.5 \cdot 24] = \frac{25}{2} \cdot 60 = 12.5 \cdot 60 = 750 \).
For \( n = 26 \): \( S_{26} = \frac{26}{2} [120 – 2.5 \cdot 25] = 13 \cdot [120 – 62.5] = 13 \cdot 57.5 = 747.5 \).
Maximum value of \( S_n \) is 750, occurring at \( n = 24 \) and \( n = 25 \) A1 A1.
[3 marks]
▶️ Answer/Explanation
(a) (i) The first \( n \) positive odd integers are 1, 3, 5, …, \( 2n-1 \). Sum in sigma notation: \( \sum_{k=1}^n (2k-1) \) A1.
Note: Award A0 for \( \sum_{n=1}^n (2n-1) \).
(ii) Method 1: The sum of the first \( n \) odd integers is \( \sum_{k=1}^n (2k-1) = 2 \sum_{k=1}^n k – \sum_{k=1}^n 1 = 2 \cdot \frac{n(n+1)}{2} – n = n(n+1) – n = n^2 \) M1 A1.
Method 2: Use the arithmetic series formula: \( S_n = \frac{n}{2} [2 \cdot 1 + (n-1) \cdot 2] = \frac{n}{2} \cdot 2n = n^2 \) M1 A1.
Method 3: Use the sum formula: \( S_n = \frac{n}{2} (u_1 + u_n) = \frac{n}{2} (1 + (2n-1)) = \frac{n}{2} \cdot 2n = n^2 \) M1 A1.
Thus, the sum is \( n^2 \) AG.
(iii) Using the result from (ii), the sum of the first 47 odd integers is \( 47^2 = 2209 \), and the sum of the first 14 odd integers is \( 14^2 = 196 \). Difference: \( 2209 – 196 = 2013 \) A1.
[4 marks]
(b) (i) For 5 points on a circle, draw a pentagon with points labeled (e.g., A, B, C, D, E). Diagonals are non-adjacent pairs: AC, AD, BE, BD, CE. Show a diagram with these 5 diagonals (circle optional) A1.
(ii) For \( n \) points, each point connects to \( n-3 \) others (excluding itself and its two adjacent points). Total connections: \( n (n-3) \) A1. Since each diagonal connects two points, divide by 2 to avoid double-counting: \( \frac{n(n-3)}{2} \) R1 R1.
(iii) Solve for the least \( n \) such that the number of diagonals exceeds 1,000,000: \( \frac{n(n-3)}{2} > 1,000,000 \) M1. Multiply through: \( n^2 – 3n > 2,000,000 \). Approximate: \( n^2 \approx 2,000,000 \), so \( n \approx \sqrt{2,000,000} \approx 1414.21 \). Solve the quadratic: \( n^2 – 3n – 2,000,000 = 0 \), roots \( n \approx 1415.7 \). Since \( n \) is an integer, try \( n = 1416 \): \( \frac{1416 \cdot 1413}{2} = 1,001,484 > 1,000,000 \); for \( n = 1415 \): \( \frac{1415 \cdot 1412}{2} = 999,070 < 1,000,000 \). Thus, \( n = 1416 \) A1 A1.
[7 marks]
(c) (i) For \( X \sim B(n, p) \), mean is \( np = 4 \), variance is \( np(1-p) = 3 \) A1. Solve: \( np(1-p) = 3 \), so \( 4(1-p) = 3 \Rightarrow 1-p = \frac{3}{4} \Rightarrow p = \frac{1}{4} \). Then \( n \cdot \frac{1}{4} = 4 \Rightarrow n = 16 \) M1 A1.
(ii) With \( X \sim B(16, 0.25) \), find \( P(X = 1 \text{ or } X = 3) = P(X = 1) + P(X = 3) \) A1.
\( P(X = 1) = \binom{16}{1} (0.25)^1 (0.75)^{15} = 16 \cdot 0.25 \cdot 0.75^{15} \approx 0.0534538 \) A1.
\( P(X = 3) = \binom{16}{3} (0.25)^3 (0.75)^{13} = \frac{16 \cdot 15 \cdot 14}{6} \cdot 0.015625 \cdot 0.75^{13} \approx 0.207876 \) A1.
Sum: \( P(X = 1) + P(X = 3) \approx 0.0534538 + 0.207876 = 0.2613298 \approx 0.261 \) (to 3 decimal places) M1 A1.
[8 marks]