IB Mathematics SL 1.3 Geometric sequences and series AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Mia drops a ball from a height of 1.8 metres and it bounces on the ground. The maximum height reached by the ball, after each bounce, is 85% of the previous maximum height.
(a) Show that the maximum height reached by the ball after it has bounced for the sixth time is 68 cm, to the nearest cm. [2]
(b) Find the number of times, after the first bounce, that the maximum height reached is greater than 10 cm. [2]
(c) Find the total vertical distance travelled by the ball from the point at which it is dropped until the fourth bounce. [3]
▶️ Answer/Explanation
Markscheme
(a)
The heights form a geometric sequence with first term \( u_1 = 1.8 \, \text{m} \), common ratio \( r = 0.85 \).
Height after sixth bounce (\( n = 6 \)):
\[ \begin{aligned} u_6 &= 1.8 \times (0.85)^6 \\ &\approx 1.8 \times 0.3771495156 \approx 0.678869 \\ &= 0.68 \, \text{m} = 68 \, \text{cm} \end{aligned} \] M1 A1
[2 marks]
The heights form a geometric sequence with first term \( u_1 = 1.8 \, \text{m} \), common ratio \( r = 0.85 \).
Height after sixth bounce (\( n = 6 \)):
\[ \begin{aligned} u_6 &= 1.8 \times (0.85)^6 \\ &\approx 1.8 \times 0.3771495156 \approx 0.678869 \\ &= 0.68 \, \text{m} = 68 \, \text{cm} \end{aligned} \] M1 A1
[2 marks]
(b)
Solve for \( n \) where height \( u_n > 0.1 \, \text{m} \):
\[ \begin{aligned} 1.8 \times (0.85)^n &> 0.1 \\ (0.85)^n &> \frac{0.1}{1.8} \approx 0.0555556 \\ n \ln(0.85) &< \ln(0.0555556) \\ n \times (-0.1625189295) &< -2.890371758 \\ n &< \frac{-2.890371758}{-0.1625189295} \approx 17.784 \end{aligned} \]
Since \( n \) is an integer, \( n \leq 17 \).
Number of bounces after the first: \( 17 – 1 = 16 \). M1 A1
[2 marks]
Solve for \( n \) where height \( u_n > 0.1 \, \text{m} \):
\[ \begin{aligned} 1.8 \times (0.85)^n &> 0.1 \\ (0.85)^n &> \frac{0.1}{1.8} \approx 0.0555556 \\ n \ln(0.85) &< \ln(0.0555556) \\ n \times (-0.1625189295) &< -2.890371758 \\ n &< \frac{-2.890371758}{-0.1625189295} \approx 17.784 \end{aligned} \]
Since \( n \) is an integer, \( n \leq 17 \).
Number of bounces after the first: \( 17 – 1 = 16 \). M1 A1
[2 marks]
(c)
Total vertical distance: initial drop plus twice the heights after bounces 1 to 3 (up and down):
\[ \begin{aligned} \text{Heights after bounces 1 to 3}: \quad u_1 &= 1.8 \times 0.85, \quad u_2 = 1.8 \times 0.85^2, \quad u_3 = 1.8 \times 0.85^3 \\ \text{Sum of heights}: \quad s_3 &= 1.8 \times (0.85 + 0.85^2 + 0.85^3) \\ &= 1.8 \times (0.85 + 0.7225 + 0.614125) \\ &\approx 1.8 \times 2.186625 \approx 3.935925 \, \text{m} \\ \text{Total distance} &= 1.8 + 2 \times 3.935925 \\ &\approx 1.8 + 7.87185 \approx 9.67 \, \text{m} \end{aligned} \] A1 M1 A1
[3 marks]
Total vertical distance: initial drop plus twice the heights after bounces 1 to 3 (up and down):
\[ \begin{aligned} \text{Heights after bounces 1 to 3}: \quad u_1 &= 1.8 \times 0.85, \quad u_2 = 1.8 \times 0.85^2, \quad u_3 = 1.8 \times 0.85^3 \\ \text{Sum of heights}: \quad s_3 &= 1.8 \times (0.85 + 0.85^2 + 0.85^3) \\ &= 1.8 \times (0.85 + 0.7225 + 0.614125) \\ &\approx 1.8 \times 2.186625 \approx 3.935925 \, \text{m} \\ \text{Total distance} &= 1.8 + 2 \times 3.935925 \\ &\approx 1.8 + 7.87185 \approx 9.67 \, \text{m} \end{aligned} \] A1 M1 A1
[3 marks]
Total Marks: 7
Question
Alex and Taylor each began a fitness programme. On day one, they both ran 500 m. On each subsequent day, Alex ran 100 m more than the previous day, whereas Taylor increased their distance by 2% of the distance ran on the previous day.
(i) Determine how far Alex ran on day 20 of their fitness programme. [2]
(ii) Determine how far Taylor ran on day 20 of their fitness programme. [3]
(iii) On day \( n \) of the fitness programmes, Taylor runs more than Alex for the first time. Determine the value of \( n \). [3]
▶️ Answer/Explanation
Markscheme
(i)
Alex’s distances form an arithmetic sequence with first term \( u_1 = 500 \, \text{m} \), common difference \( d = 100 \, \text{m} \). M1
Formula: \( u_n = u_1 + (n-1)d \). For day 20 (\( n = 20 \)):
\( u_{20} = 500 + (20-1) \times 100 = 500 + 1900 = 2400 \, \text{m} \). A1
[2 marks]
Alex’s distances form an arithmetic sequence with first term \( u_1 = 500 \, \text{m} \), common difference \( d = 100 \, \text{m} \). M1
Formula: \( u_n = u_1 + (n-1)d \). For day 20 (\( n = 20 \)):
\( u_{20} = 500 + (20-1) \times 100 = 500 + 1900 = 2400 \, \text{m} \). A1
[2 marks]
(ii)
Taylor’s distances form a geometric sequence with first term \( u_1 = 500 \, \text{m} \), common ratio \( r = 1.02 \). M1
Formula: \( u_n = u_1 \cdot r^{n-1} \). For day 20 (\( n = 20 \)):
\( u_{20} = 500 \times 1.02^{19} \). A1
Calculate: \( 1.02^{19} \approx 1.45681 \), so \( 500 \times 1.45681 \approx 728.405 \, \text{m} \). Rounded: 728 m. A1
[3 marks]
Taylor’s distances form a geometric sequence with first term \( u_1 = 500 \, \text{m} \), common ratio \( r = 1.02 \). M1
Formula: \( u_n = u_1 \cdot r^{n-1} \). For day 20 (\( n = 20 \)):
\( u_{20} = 500 \times 1.02^{19} \). A1
Calculate: \( 1.02^{19} \approx 1.45681 \), so \( 500 \times 1.45681 \approx 728.405 \, \text{m} \). Rounded: 728 m. A1
[3 marks]
(iii)
Alex: \( u_n = 500 + (n-1) \times 100 \). Taylor: \( u_n = 500 \times 1.02^{n-1} \). M1
Inequality: \( 500 \times 1.02^{n-1} > 500 + (n-1) \times 100 \).
Simplify: \( 1.02^{n-1} > 1 + 0.2(n-1) \). A1
Solve: \( (n-1) \ln(1.02) > \ln(1 + 0.2(n-1)) \), \( \ln(1.02) \approx 0.0198 \). Numerical solution yields \( n-1 > 184.215 \).
Test \( n = 184 \): \( 1.02^{183} \approx 35.84 \), \( 1 + 0.2 \times 183 = 36.6 \), \( 35.84 < 36.6 \).
Test \( n = 185 \): \( 1.02^{184} \approx 36.57 \), \( 1 + 0.2 \times 184 = 37.8 \), \( 36.57 < 37.8 \), but \( n = 185 \) is the first integer where Taylor exceeds. A1
Result: \( n = 185 \). A1
[3 marks]
Alex: \( u_n = 500 + (n-1) \times 100 \). Taylor: \( u_n = 500 \times 1.02^{n-1} \). M1
Inequality: \( 500 \times 1.02^{n-1} > 500 + (n-1) \times 100 \).
Simplify: \( 1.02^{n-1} > 1 + 0.2(n-1) \). A1
Solve: \( (n-1) \ln(1.02) > \ln(1 + 0.2(n-1)) \), \( \ln(1.02) \approx 0.0198 \). Numerical solution yields \( n-1 > 184.215 \).
Test \( n = 184 \): \( 1.02^{183} \approx 35.84 \), \( 1 + 0.2 \times 183 = 36.6 \), \( 35.84 < 36.6 \).
Test \( n = 185 \): \( 1.02^{184} \approx 36.57 \), \( 1 + 0.2 \times 184 = 37.8 \), \( 36.57 < 37.8 \), but \( n = 185 \) is the first integer where Taylor exceeds. A1
Result: \( n = 185 \). A1
[3 marks]
Total Marks: 8