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Date November 2012 Marks available 6 Reference code 12N.2.hl.TZ0.5
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 5 Adapted from N/A

Question

A metal rod 1 metre long is cut into 10 pieces, the lengths of which form a geometric sequence. The length of the longest piece is 8 times the length of the shortest piece. Find, to the nearest millimetre, the length of the shortest piece.

Markscheme

the pieces have lengths \(a,{\text{ }}ar,{\text{ ..., }}a{r^9}\)     (M1)

\(8a = a{r^9}{\text{ }}({\text{or }}8 = {r^9})\)     A1

\(r = \sqrt[9]{8} = 1.259922...\)     A1

\(a\frac{{{r^{10}} - 1}}{{r - 1}} = 1\,\,\,\,\,\left( {{\text{or }}a\frac{{{r^{10}} - 1}}{{r - 1}} = 1000} \right)\)     M1

\(a = \frac{{r - 1}}{{{r^{10}} - 1}} = 0.0286...\,\,\,\,\,\left( {{\text{or }}a = \frac{{r - 1}}{{{r^{10}} - 1}} = 28.6...} \right)\)     (A1)

a = 29 mm (accept 0.029 m or any correct answer regardless the units)     A1

[6 marks]

Examiners report

This question was generally well done by most candidates. Some candidates recurred to a diagram to comprehend the nature of the problem but a few thought it was an arithmetic sequence.

A surprising number of candidates missed earning the final A1 mark because they did not read the question instructions fully and missed the accuracy instruction to give the answer correct to the nearest mm.

Syllabus sections

Topic 1 - Core: 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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