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Date	November Example questions	Marks available	4	Reference code	EXN.1.SL.TZ0.4
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Prove	Question number	4	Adapted from	N/A

Question

The first three terms of an arithmetic sequence are $u_{1}, 5 u_{1} - 8$ and $3 u_{1} + 8$ .

Show that $u_{1} = 4$ .

[2]

a.

Prove that the sum of the first $n$ terms of this arithmetic sequence is a square number.

[4]

b.

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.

EITHER

uses $u_{2} - u_{1} = u_{3} - u_{2}$ (M1)

$(5 u_{1} - 8) - u_{1} = (3 u_{1} + 8) - (5 u_{1} - 8)$

$6 u_{1} = 24$ A1

OR

uses $u_{2} = \frac{u_{1} + u_{3}}{2}$ (M1)

$5 u_{1} - 8 = \frac{u_{1} + (3 u_{1} + 8)}{2}$

$3 u_{1} = 12$ A1

THEN

so $u_{1} = 4$ AG

[2 marks]

a.

$d = 8$ (A1)

uses $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ M1

$S_{n} = \frac{n}{2} (8 + 8 (n - 1))$ A1

$= 4 n^{2}$

$= {(2 n)}^{2}$ A1

Note: The final A1 can be awarded for clearly explaining that $4 n^{2}$ is a square number.

so sum of the first $n$ terms is a square number AG

[4 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

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

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