Edexcel A level Maths (9MA0) · Pure mathematics › Sequences and series
Practise Modelling with sequences and series. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A factory makes 1300 bicycles in week 1. The number made increases by 40 each week.
(a) Find the number of bicycles made in week 19.[2]
(b) Find the total number of bicycles made in the first 15 weeks.[2]
Show the answer and mark scheme
(a)Answer: 2020
M1 for \(1300 + 18 \times 40\)
A1 for 2020
Worked solution: \(1300 + (19 - 1) \times 40 = 2020\), so the number of bicycles made in week 19 is 2020.
(b)Answer: 23,700
M1 for \(\frac{15}{2}\left[2 \times 1300 + 14 \times 40\right]\)
A1 for 23,700
Worked solution: \(S_{15} = \frac{15}{2}\left[2 \times 1300 + 14 \times 40\right] = 23{,}700\), so the total number of bicycles made in the first 15 weeks is 23,700.
Question 2Medium7 marks
A factory makes 1500 bicycles in week 1. The number made increases by 50 each week.
(a) Find the number of bicycles made in week 13.[2]
(b) Find the week in which the total number of bicycles made since week 1 first exceeds 40,000.[4]
(c) Give one reason why this model may not be realistic in the long term.[1]
Show the answer and mark scheme
(a)Answer: 2100
M1 for \(1500 + 12 \times 50\)
A1 for 2100
Worked solution: \(1500 + (13 - 1) \times 50 = 2100\), so the number of bicycles made in week 13 is 2100.
A1 for a correct 3-term quadratic, e.g. \(n^{2} + 59n - 1600 \gt 0\)
M1 for solving their quadratic to find the positive critical value, \(n = 20.20\)
A1 for week 21
Worked solution: \(\frac{n}{2}\left[3000 + 50(n - 1)\right] \gt 40{,}000 \Rightarrow n^{2} + 59n - 1600 \gt 0\). The positive root is \(n = 20.202\), so the total first exceeds 40,000 in week 21. (Check: after 20 weeks the total is 39,500; after 21 it is 42,000.)
(c)Answer: E.g. production cannot increase indefinitely (limited factory capacity, workers or demand)
B1 for a sensible reason in context, e.g. production cannot increase indefinitely (limited factory capacity, workers or demand)
Worked solution: For example, production cannot increase indefinitely (limited factory capacity, workers or demand).
Question 3Hard10 marks
A ball is dropped from a height of 3 m onto horizontal ground. Each time it hits the ground it bounces back up to 70% of the height from which it last fell. The maximum heights reached after successive bounces are modelled as a geometric sequence.
(a) Find the maximum height reached by the ball after its 4th bounce, giving your answer in metres to 3 significant figures.[2]
(b) Show that the total distance travelled by the ball, from when it is dropped until it hits the ground for the 6th time, is 14.6 m to 3 significant figures.[4]
(c) Find the total distance that the ball travels according to the model.[3]
(d) Give one criticism of the model.[1]
Show the answer and mark scheme
(a)Answer: 0.720 m
M1 for \(3 \times 0.7^{4}\)
A1 for 0.720 (awrt 0.720)
Worked solution: After the first bounce the ball rises to \(3 \times 0.7\) m, so after the 4th bounce it rises to \(3 \times 0.7^{4} = 0.7203\) m, i.e. 0.720 m (3 s.f.).
(b)Answer: Shown
B1 for identifying the heights after the first five bounces, \(2.1, 1.47, 1.029, \ldots\), as a geometric series with first term 2.1 and common ratio 0.7
M1 for recognising that each of these heights is travelled twice (up and down), in addition to the initial drop of 3 m
A1* for \(3 + 11.647\ldots = 14.647\ldots = 14.6\) m (cso)
Worked solution: Between the first and the 6th impacts the ball makes 5 bounces, rising to \(2.1, 1.47, 1.029, 0.7203, 0.50421\) m, and each height is travelled up and then down. \(\text{Total} = 3 + 2 \times \frac{2.1(1 - 0.7^{5})}{1 - 0.7} = 3 + 2 \times 5.82351 = 14.647\ldots\), which is 14.6 m to 3 significant figures.
(c)Answer: 17 m
M1 for using a sum to infinity for the bounces, \(\frac{2.1}{1 - 0.7} = 7\)
M1 for \(3 + 2 \times 7\)
A1 for 17 (m)
Worked solution: The heights after the bounces are \(2.1, 1.47, 1.029, \ldots\), a geometric series with sum to infinity \(\frac{2.1}{1 - 0.7} = 7\). Each is travelled twice (up and down), so the total is \(3 + 2 \times 7 = 17\) m.
(d)Answer: e.g. the model predicts infinitely many bounces
B1 for a sensible criticism, e.g. the model predicts that the ball bounces infinitely many times, whereas a real ball stops after a finite number of bounces; air resistance is ignored; the ball will not bounce to exactly 70% each time
Worked solution: For example, the model predicts that the ball never stops bouncing, but in reality the bounces die out after a finite number, because the proportion of height regained is not constant for very small bounces.