IB Mathematics SL 1.2 Arithmetic Sequences & Series AA SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)
Recognize arithmetic sequence with first term \( a = 5 \), common difference \( d = 4 \) (M1)
Use term formula: \( u_n = a + (n-1)d = 5 + (n-1) \cdot 4 = 4n + 1 \) (A1)
Solve: \( 4n + 1 = 801 \implies 4n = 800 \implies n = 200 \) (A1) (AG)
[3 marks]
Part (b)
Use sum formula: \( S_n = \frac{n}{2} \left[ a + u_n \right] \) (M1)
Substitute \( n = 200 \), \( a = 5 \), \( u_{200} = 801 \): \( S_{200} = \frac{200}{2} \cdot (5 + 801) = 100 \cdot 806 \) (A1)
\( S_{200} = 80600 \) (A1) (N2)
Answer: 80,600 [3 marks]
▶️ Answer/Explanation
Part (a)
Use term formula: \( u_n = a + (n-1)d \) (M1)
Substitute: \( u_k = 60 + (k-1)(-2.5) = 0 \implies 60 – 2.5(k-1) = 0 \implies 2.5(k-1) = 60 \implies k-1 = 24 \implies k = 25 \) (A1) (N2)
Answer: \( k = 25 \) [2 marks]
Part (b)
Recognize maximum \( S_n \) at \( n = k = 25 \) (when \( u_{25} = 0 \)) (M1)
Use sum formula: \( S_n = \frac{n}{2} (a + u_n) \) or \( S_n = \frac{n}{2} [2a + (n-1)d] \) (A1)
Substitute \( n = 25 \), \( a = 60 \), \( u_{25} = 0 \): \( S_{25} = \frac{25}{2} (60 + 0) = 12.5 \cdot 60 = 750 \) (A1) (N2)
Answer: \( S_{25} = 750 \) [3 marks]
▶️ Answer/Explanation
Part (a)
Common difference from \( u_n = 5 + 2n \): \( u_{n+1} – u_n = (5 + 2(n+1)) – (5 + 2n) = 2 \) (A1)
Answer: \( d = 2 \) (N1) [1 mark]
Part (b)
(i) Solve: \( u_n = 5 + 2n = 115 \implies 2n = 110 \implies n = 55 \) (A1, A1) (N2)
Answer: \( n = 55 \) [2 marks]
(ii) First term: \( u_1 = 5 + 2 \cdot 1 = 7 \) (A1)
Use sum formula: \( S_n = \frac{n}{2} (u_1 + u_n) \) or \( S_n = \frac{n}{2} [2u_1 + (n-1)d] \) (A1)
Substitute \( n = 55 \), \( u_1 = 7 \), \( u_{55} = 115 \): \( S_{55} = \frac{55}{2} (7 + 115) = \frac{55}{2} \cdot 122 = 55 \cdot 61 = 3355 \) (A1) (N3)
Answer: \( S_{55} = 3355 \) [3 marks]