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Date	November Example question	Marks available	2	Reference code	EXN.1.SL.TZ0.12
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Find	Question number	12	Adapted from	N/A

Question

A disc is divided into $9$ sectors, number $1$ to $9$ . The angles at the centre of each of the sectors $u_{n}$ form an arithmetic sequence, with $u_{1}$ being the largest angle.

It is given that $u_{9} = \frac{1}{3} u_{1}$ .

Write down the value of $Σ_{i = 1}^{9} u_{i}$ .

[1]

a.

Find the value of $u_{1}$ .

[4]

b.

A game is played in which the arrow attached to the centre of the disc is spun and the sector in which the arrow stops is noted. If the arrow stops in sector $1$ the player wins $10$ points, otherwise they lose $2$ points.

Let $X$ be the number of points won

Find $E (X)$ .

[2]

c.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

$360 °$ A1

[1 mark]

a.

EITHER

$360 = \frac{9}{2} (u_{1} + u_{9})$ M1

$360 = \frac{9}{2} (u_{1} + \frac{1}{3} u_{1}) = 6 u_{1}$ M1A1

OR

$360 = \frac{9}{2} (2 u_{1} + 8 d)$ M1

$u_{9} = \frac{1}{3} u_{1} = u_{1} + 8 d \Rightarrow u_{1} = - 12 d$ M1

Substitute this value $360 = \frac{9}{2} (2 u_{1} - 8 \times \frac{u_{1}}{12}) (= \frac{9}{2} \times \frac{4}{3} u_{1} = 6 u_{1})$ A1

THEN

$u_{1} = 60 °$ A1

[4 marks]

b.

$E (X) = 10 \times \frac{60}{360} - 2 \times \frac{300}{360} = 0$ M1A1

[2 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

Syllabus sections

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

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