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Date May 2016 Marks available 3 Reference code 16M.1.sl.TZ1.4
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 4 Adapted from N/A

Question

Consider the following sequence of figures.

M16/5/MATME/SP1/ENG/TZ1/04

Figure 1 contains 5 line segments.

Given that Figure \(n\) contains 801 line segments, show that \(n = 200\).

[3]
a.

Find the total number of line segments in the first 200 figures.

[3]
b.

Markscheme

recognizing that it is an arithmetic sequence     (M1)

eg\(\,\,\,\,\,\)\(5,{\text{ }}5 + 4,{\text{ }}5 + 4 + 4,{\text{ }} \ldots ,{\text{ }}d = 4,{\text{ }}{u_n} = {u_1} + (n - 1)d,{\text{ }}4n + 1\)

correct equation     A1

eg\(\,\,\,\,\,\)\(5 + 4(n - 1) = 801\)

correct working (do not accept substituting \(n = 200\))     A1

eg\(\,\,\,\,\,\)\(4n - 4 = 796,{\text{ }}n - 1 = \frac{{796}}{4}\)

\(n = 200\)    AG     N0

[3 marks]

a.

recognition of sum     (M1)

eg\(\,\,\,\,\,\)\({S_{200}},{\text{ }}{u_1} + {u_2} +  \ldots  + {u_{200}},{\text{ }}5 + 9 + 13 +  \ldots  + 801\)

correct working for AP     (A1)

eg\(\,\,\,\,\,\)\(\frac{{200}}{2}(5 + 801),{\text{ }}\frac{{200}}{2}{\text{ }}\left( {2(5) + 199(4)} \right)\)

\(80\,600\)     A1     N2

[3 marks]

b.

Examiners report

Most candidates recognized that the series was arithmetic but many worked backwards using \(n = 200\) rather than creating and solving an equation of their own to produce the given answer.

a.

Almost all students answered (b) correctly.

b.

Syllabus sections

Topic 1 - Algebra » 1.1 » Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.
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