IBDP Maths SL 1.3 Geometric sequences and series AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) The height after the \( n \)-th bounce forms a geometric sequence with first term \( u_1 = 4 \) and common ratio \( r = 0.95 \). The height after the fourth bounce is \( u_5 = 4 \times 0.95^4 \) (since the first bounce is \( n=1 \)) M1.
Calculate: \( 0.95^4 = 0.81450625 \), so \( 4 \times 0.81450625 = 3.258025 \approx 3.26 \) metres (to 3 significant figures) A1.
[2 marks]
(b) The height after \( n \) bounces is \( u_{n+1} = 4 \times 0.95^n \). Solve for when the height falls below 1 metre: \( 4 \times 0.95^n < 1 \) M1.
Simplify: \( 0.95^n < 0.25 \).
Take logarithms: \( n \ln 0.95 < \ln 0.25 \). Since \( \ln 0.95 < 0 \), reverse the inequality: \( n > \frac{\ln 0.25}{\ln 0.95} \approx 27.0296 \) A1.
Since \( n \) is an integer, the smallest \( n \) for which \( 0.95^n < 0.25 \) is \( n = 28 \). Thus, the ball bounces 28 times before the height is less than 1 metre A1.
Note: Do not penalize improper use of inequalities.
[3 marks]
(c) Method 1: The total distance includes the initial drop (4 metres) plus the distance for each bounce (up and down, each at the same height). After the first drop, each bounce \( n \geq 1 \) contributes \( 2 \times 4 \times 0.95^{n-1} \). This forms a geometric series with first term \( 4 \times 0.95 \) and common ratio 0.95 M1.
Sum to infinity: \( \frac{4 \times 0.95}{1 – 0.95} = \frac{3.8}{0.05} = 76 \). Double this for up and down, and add the initial drop: \( 2 \times 76 + 4 = 152 + 4 = 156 \) metres M1 A1.
Method 2: Consider the total distance as the initial drop (4 metres) plus all bounces (up and down). The series for one direction (after the initial drop) is \( 4 + 4 \times 0.95 + 4 \times 0.95^2 + \dots \), with first term 4 and common ratio 0.95 M1.
Sum to infinity: \( \frac{4}{1 – 0.95} = \frac{4}{0.05} = 80 \). Double this for up and down, subtract the initial drop (counted only once): \( 2 \times 80 – 4 = 160 – 4 = 156 \) metres M1 A1.
Note: If candidates use \( n-1 \) instead of \( n \) for bounces, penalize in part (a) and treat as follow-through in parts (b) and (c).
[3 marks]
▶️ Answer/Explanation
(a) Initial value = $35,000. A 15% decrease means the car retains 85% of its value: multiplier = 0.85 M1.
Value after first year = \( 35,000 \times 0.85 = 29,750 \) dollars A1.
[2 marks]
(b) After the first year, value = $29,750. Each subsequent year, the value decreases by 11%, so multiplier = 0.89. For 10 years total, 9 years have 11% decrease M1.
Value after 10 years = \( 29,750 \times 0.89^9 \approx 29,750 \times 0.3513 \approx 10,451.17 \) A1.
To the nearest dollar: $10,451 A1.
[3 marks]
(c) Find the least \( n \) such that the value is less than 10% of $35,000, i.e., $3,500. Value after \( n \) years = \( 35,000 \times 0.85 \times 0.89^{n-1} < 3,500 \) M1.
Simplify: \( 29,750 \times 0.89^{n-1} < 3,500 \Rightarrow 0.89^{n-1} < \frac{3,500}{29,750} \approx 0.1176 \).
Solve: \( \ln(0.89^{n-1}) < \ln(0.1176) \Rightarrow (n-1) \ln(0.89) < \ln(0.1176) \). Since \( \ln(0.89) \approx -0.1165 \), \( \ln(0.1176) \approx -2.1400 \), we get \( n – 1 > \frac{-2.1400}{-0.1165} \approx 18.37 \Rightarrow n > 19.37 \) A1.
Least integer \( n = 20 \) A1.
[3 marks]