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IB Mathematics SL 1.2 Arithmetic sequences and series AI SL Paper 2 - Exam Style Questions - New Syllabus

IB DP Maths AI SL Paper 2 - All Topics

Question 

A new Auditorium was built with \(14\) seats in the first row. Each subsequent row of the hall has two more seats than the previous row. The hall has a total of \(20\) rows.
(a) Find:
(i) the number of seats in the last row.
(ii) the total number of seats in the Auditorium. [5]
The Auditorium opened in 2019. The average number of visitors per concert during that year was \(584\). In 2020, the average number of visitors per concert increased by \(1.2\%\).
(b) Find the average number of visitors per concert in 2020. [2]
The concert organizers use this data to model future numbers of visitors. It is assumed that the average number of visitors per concert will continue to increase each year by \(1.2\%\).
(c) Determine the first year in which this model predicts the average number of visitors per concert will exceed the total seating capacity of the Auditorium. [5]
It is assumed that the Auditorium will host \(50\) concerts each year.
(d) Use the average number of visitors per concert per year to predict the total number of people expected to attend the Auditorium from when it opens until the end of 2025. [4]
▶️ Answer / Explanation
Markscheme with detailed working

(a)

Let the number of seats in row \(n\) be \(a_n\). This is an arithmetic sequence with first term \(a_1=14\) and common difference \(d=2\).

(i) Last row is \(n=20\): \[ a_{20} = a_1 + (20-1)d = 14 + 19\times 2 = 14 + 38 = \boxed{52}. \] M1 M1 A1
(ii) Total seats \(S_{20} = \dfrac{20}{2}\,(a_1+a_{20}) = 10\,(14+52)=10\times 66=\boxed{660}.\; \text{(Arithmetic series formula)}\)M1 A1
[5 marks]

(b)

Increase by \(1.2\%\) \(\Rightarrow\) multiply by \(1.012\): \[ 584\times 1.012 = 584 + 0.012\times 584 = 584 + 7.008 = 591.008. \] Average \(\approx \boxed{591}\) (exact \(=591.008\)). M1 A1

[2 marks]

(c)

Model the average visitors per concert in year \(n\) (with \(n=0\) for 2019) by the geometric sequence \[ v_n = 584\,(1.012)^n. \] Find smallest \(n\) with \(v_n > 660\): \[ 584(1.012)^n > 660 \;\;\Longleftrightarrow\;\; (1.012)^n > \frac{660}{584}=1.130136986\ldots \] Taking logs: \[ n > \frac{\ln(660/584)}{\ln(1.012)} \approx \frac{\ln(1.130136986)}{\ln(1.012)} \approx \frac{0.1222}{0.0119} \approx 10.256. \] Hence the first integer \(n\) is \(11\). Year \(= 2019 + 11 = \boxed{2030}.\) M1 M1 A1 A1 A1

[5 marks]

(d)

Years from opening (2019) to end of 2025 inclusive are \(7\) years: \(n=0,1,\dots,6\).

Sum of yearly averages per concert: \[ \sum_{k=0}^{6} 584(1.012)^k = 584\;\frac{(1.012)^7-1}{1.012-1} = 584\;\frac{1.086\;(\text{approx})-1}{0.012} \approx 584\cdot 7.2554 \approx 4238.1469. \] With \(50\) concerts each year, total people \[ \text{Total} = 50\times 4238.1469 \approx \boxed{212\,000}\;\text{(exact }211\,907.3\ldots\text{)}. \] A1 M1 M1 A1
[4 marks]
Total: 16 marks
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