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Arithmetic Sequences and Series · Paper 3 — Free Maths AI HL Practice Questions

1ChallengeLAQSum of an arithmetic sequence6 marksPaper 3~9 min
A farmer constructs a series of trapezoidal irrigation channels. The cross-section of each channel is a trapezium with a constant bottom width of 1.0 m1.0\,\text{m} and a constant depth of 0.5 m0.5\,\text{m}. The first channel has a top width of 2.0 m2.0\,\text{m}, and each subsequent channel has a top width 0.4 m0.4\,\text{m} wider than the previous one.
(a)
Prove that the cross-sectional area AnA_n of the nnth channel satisfies An=0.5(3+0.4(n−1)) m2.A_n = 0.5\bigl(3 + 0.4(n-1)\bigr)\,\text{m}^2. [3 marks]
(b)
The cross-sectional areas A1,A2,…A_1, A_2, \ldots form an arithmetic sequence. (i) Write down the first term aa and common difference dd of this arithmetic sequence. [2 marks]
(ii) Find the total cross-sectional area of the first 2020 channels. [2 marks]
(iii) The farmer wants to build channels until the total cross-sectional area first exceeds 200 m2200\,\text{m}^2. Find the minimum number of channels needed. [3 marks]
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2ChallengeLAQApplications of arithmetic sequences in real-world contexts9 marksPaper 3~14 min
A construction company is building a high-rise tower. The first floor has a height of 5.0 m5.0\,\text{m}. Each subsequent floor has a height 0.10 m0.10\,\text{m} less than the floor below it. The building has 4040 floors in total.
(a)
Show that the height of the nnth floor is given by un=5.1−0.1nu_n = 5.1 - 0.1n metres, and state the height of the 40th floor. [2 marks]
(b)
Calculate the total height of the building from ground level to the top of the 40th floor. [3 marks]
(c)
The building must comply with a regulation stating that the average floor height must be at least 3.0 m3.0\,\text{m}. Evaluate whether this building meets the regulation. Justify your conclusion. [2 marks]
(d)
A second tower has the same number of floors. Its floor heights also form an arithmetic sequence, with a first floor height of 5.0 m5.0\,\text{m} and a 40th floor height of 3.5 m3.5\,\text{m}. Show that the common difference of this sequence differs from that of the first tower. [1 mark] (e) Determine which tower has the greater total height, and calculate the difference in total heights between the two towers. [2 marks]
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Solutions

3ChallengeLAQApplications of arithmetic sequences in real-world contexts11 marksPaper 3~17 min
A pharmaceutical company is testing a new drug. In the first clinical trial, 120120 patients receive the drug. The number of patients in each subsequent trial increases by a constant amount. The company runs 1515 trials in total. In the 1515th trial, 330330 patients participate.
(a)
Show that the number of patients across the 1515 trials forms an arithmetic sequence with first term u1=120u_1 = 120, 1515th term u15=330u_{15} = 330, and common difference d=15d = 15. [2 marks]
(b)
Calculate the number of patients in the 88th trial. [2 marks]
(c)
Calculate the total number of patients across all 1515 trials. [3 marks]
(d)
The company's budget allows for a maximum of 30003000 patients across all trials. Evaluate whether the total number of patients exceeds this budget. Justify your conclusion. [2 marks] (e) In a separate study, a different drug is trialled over nn trials. The number of patients in each trial again increases by a constant amount. The total number of patients across all nn trials is 35003500, the last trial has 240240 patients, and the common difference is 1010. Determine the value of nn. [4 marks]
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4ChallengeLAQDefinition and general term of arithmetic sequences6 marksPaper 3~9 min
A manufacturing company produces cylindrical metal rods. The lengths of the rods, in millimetres, form an arithmetic sequence with first term u1u_1 and common difference dd.
(a)
Show that the length of the nnth rod is given by un=u1+(n−1)du_n = u_1 + (n-1)d, by considering the recurrence relation un=un−1+du_n = u_{n-1} + d. [3 marks]
(b)
The 5th rod has length 170 mm170\,\text{mm} and the 12th rod has length 198 mm198\,\text{mm}. (i) Find the value of dd and the value of u1u_1. [4]
(ii) Find the total length, in mm, of the first 15 rods placed end to end. [2]
(iii) The company claims that the total length of the first nn rods exceeds 10 m10\,\text{m} for the first time when n=18n = 18. Determine whether this claim is correct, showing all working. [3 marks]

Solutions

5ChallengeLAQDefinition and general term of arithmetic sequences7 marksPaper 3~11 min
A company manufactures storage tanks. The volume of the smallest tank is 500500 litres. Each subsequent tank has a volume that is dd litres larger than the previous one, forming an arithmetic sequence. The total volume of the first 1212 tanks is 93009300 litres.
(a)
Show that d=50d = 50. [3 marks]
(b)
The company redesigns the range so that, instead of a constant difference, the volume increases by a fixed percentage p%p\% each time. The smallest tank remains 500500 litres, and the volume of the 1212th tank equals the volume of the 1212th tank in the original arithmetic design. (i) Find the value of pp, correct to 33 significant figures. [3 marks]
(ii) Calculate the total volume of the first 1212 tanks in the percentage-increase design, and hence determine whether it is greater than, less than, or equal to 93009300 litres. [4 marks]

Solutions

6ChallengeLAQSum of an arithmetic sequence10 marksPaper 3~15 min
A company is developing a new energy storage system. During testing, the amount of energy (in megawatt-hours) stored during each full charge-discharge cycle follows an arithmetic sequence. In the first cycle the system stores u1=120 MWhu_1 = 120\,\text{MWh}. In the fifth cycle it stores u5=96 MWhu_5 = 96\,\text{MWh}. The total energy stored over the first nn cycles is denoted SnS_n.
(a)
Prove that the sum of the first nn terms of an arithmetic sequence with first term aa and common difference dd is Sn=n2(2a+(n−1)d).S_n = \frac{n}{2}\bigl(2a + (n-1)d\bigr). [3 marks]
(b)
Determine the common difference dd and hence show that Sn=3n(41−n).S_n = 3n(41 - n). [3 marks]
(c)
The system is retired immediately after the cycle in which the cumulative total energy stored first exceeds 1000 MWh1000\,\text{MWh}. Determine the cycle number at which retirement occurs, justifying your answer fully. [4 marks]

Solutions

7ChallengeLAQSum of an arithmetic sequence9 marksPaper 3~14 min
A company designs a tiered bonus structure for its employees. In the first year, an employee receives a bonus of 500500 dollars. Each subsequent year, the bonus increases by a fixed amount of dd dollars compared to the previous year. The total bonus received over nn years is denoted SnS_n.
(a)
Prove that the sum of the first nn terms of an arithmetic sequence with first term aa and common difference dd is Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}\left[2a + (n-1)d\right] [4 marks]
(b)
The total bonus received over the first 1010 years is 72507250 dollars. (i) Show that d=50d = 50. [3 marks]
(ii) Hence, find the bonus received in the 1515th year. [2 marks]
(c)
The company requires the total bonus paid over the first nn years to be at least 50 00050\,000 dollars. Determine the minimum value of nn for which this possible. [3 marks]

Solutions

8ChallengeLAQSum of an arithmetic sequence10 marksPaper 3~15 min
A biologist is studying the growth of a bacterial colony. The number of bacteria (in thousands) counted at the start of each hour forms an arithmetic sequence with first term aa and common difference dd. At the start of hour 1 (t=1t = 1), there are 5050 thousand bacteria. At the start of hour 4 (t=4t = 4), there are 6868 thousand bacteria.
(a)
Show that the total number of bacteria (in thousands) counted from the start of hour 1 to the start of hour nn is given by Tn=n2(2a+(n−1)d)T_n = \frac{n}{2}\bigl(2a + (n-1)d\bigr) by writing TnT_n as a sum in two orders and adding the results. [3 marks]
(b)
Find the common difference dd, and hence show that Tn=n(47+3n).T_n = n(47 + 3n). [3 marks]
(c)
The biologist requires a cumulative total of at least 20002000 thousand bacteria. Determine the smallest integer value of nn for which Tn≥2000T_n \geq 2000. [2 marks]
(d)
The arithmetic model predicts that the number of bacteria at the start of hour nn is un=50+6(n−1)u_n = 50 + 6(n-1). The biologist observes that, after hour 15, the growth rate begins to slow because of nutrient depletion. Evaluate whether the value of nn found in part (c) is likely to be an underestimate or an overestimate of the true number of hours required, justifying your answer. [2 marks]

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