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Date	May 2021	Marks available	2	Reference code	21M.2.SL.TZ2.3
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	3	Adapted from	N/A

Question

A new concert hall 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.

Find:

The concert hall 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 $1.2 %$ .

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 $1.2 %$ .

the number of seats in the last row.

[3]

a.i.

the total number of seats in the concert hall.

[2]

a.ii.

Find the average number of visitors per concert in $2020$ .

[2]

Determine the first year in which this model predicts the average number of visitors per concert will exceed the total seating capacity of the concert hall.

[5]

It is assumed that the concert hall will host $50$ concerts each year.

Use the average number of visitors per concert per year to predict the total number of people expected to attend the concert hall from when it opens until the end of $2025$ .

[4]

Markscheme

recognition of arithmetic sequence with common difference $2$ (M1)

use of arithmetic sequence formula (M1)

$14 + 2 (20 - 1)$

$52$ A1

[3 marks]

a.i.

use of arithmetic series formula (M1)

$\frac{14 + 52}{2} \times 20$

$660$ A1

[2 marks]

a.ii.

$584 + (584 \times 0.012)$ OR $584 \times {(1.012)}^{1}$ (M1)

$591 (591.008)$ A1

Note: Award M0A0 if incorrect $r$ used in part (b), and FT with their $r$ in parts (c) and (d).

[2 marks]

recognition of geometric sequence (M1)

equating their $n$ th geometric sequence term to their $660$ (M1)

Note: Accept inequality.

METHOD 1

EITHER

$600 = 584 \times {(1.012)}^{x - 1}$ A1

$(x - 1 =) 10.3 (10.2559 \dots)$

$x = 11.3 (11.2559 \dots)$ A1

$2030$ A1

$600 = 584 \times {(1.012)}^{x}$ A1

$x = 10.3 (10.2559 \dots)$ A1

$2030$ A1

METHOD 2

11^th term $658 (657.987 \dots)$ (M1)A1

12^th term $666 (666.883 \dots)$ (M1)A1

$2030$ A1

Note: The last mark can be awarded if both their 11^th and 12^th correct terms are seen.

[5 marks]

$7$ seen (A1)

EITHER

$584 (\frac{1 . 012^{7} - 1}{1.012 - 1})$ (M1)

multiplying their sum by $50$ (M1)

sum of the number of visitors for their $r$ and their seven years (M1)

multiplying their sum by $50$ (M1)

$29 200 (\frac{1 . 012^{7} - 1}{1.012 - 1})$ (M1)(M1)

THEN

$212000 (211907.3 \dots)$ A1

Note: Follow though from their $r$ from part (b).

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.2—Arithmetic sequences and series

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21N.2.SL.TZ0.2c:
Write down an expression for $v_{n}$ .
21N.2.SL.TZ0.2d:
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21N.2.SL.TZ0.2e:
Find $k$ .

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Topic 1—Number and algebra

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