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Date	May 2021	Marks available	3	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

An arithmetic sequence has first term $60$ $60$ and common difference $- 2.5$ $- 2.5$ .

Given that the $k$ $k$ th term of the sequence is zero, find the value of $k$ $k$ .

[2]

a.

Let $S_{n}$ $S_{n}$ denote the sum of the first $n$ $n$ terms of the sequence.

Find the maximum value of $S_{n}$ $S_{n}$ .

[3]

b.

Markscheme

attempt to use $u_{1} + (n - 1) d = 0$ $u_{1} + (n - 1) d = 0$ (M1)

$60 - 2.5 (k - 1) = 0$ $60 - 2.5 (k - 1) = 0$

$k = 25$ $k = 25$ A1

[2 marks]

a.

METHOD 1

attempting to express $S_{n}$ $S_{n}$ in terms of $n$ $n$ (M1)

use of a graph or a table to attempt to find the maximum sum (M1)

$= 750$ $= 750$ A1

METHOD 2

EITHER

recognizing maximum occurs at $n = 25$ $n = 25$ (M1)

$S_{25} = \frac{25}{2} (60 + 0), S_{25} = \frac{25}{2} (2 \times 60 + 24 \times - 2.5)$ $S_{25} = \frac{25}{2} (60 + 0), S_{25} = \frac{25}{2} (2 \times 60 + 24 \times - 2.5)$ (A1)

OR

attempting to calculate $S_{24}$ $S_{24}$ (M1)

$S_{24} = \frac{24}{2} (2 \times 60 + 23 \times - 2.5)$ $S_{24} = \frac{24}{2} (2 \times 60 + 23 \times - 2.5)$ (A1)

THEN

$= 750$ $= 750$ A1

[3 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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