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Date May 2008 Marks available 3 Reference code 08M.1.sl.TZ1.3
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 3 Adapted from N/A

Question

Consider the arithmetic sequence 25811, .

Find u101 .

[3]
a.

Consider the arithmetic sequence 25811, .

Find the value of n so that un=152 .

[3]
b.

Markscheme

d=3     (A1)

evidence of substitution into un=a+(n1)d     (M1)

e.g. u101=2+100×3

u101=302     A1     N3

[3 marks]

a.

correct approach     (M1)

e.g. 152=2+(n1)×3

correct simplification     (A1)

e.g. 150=(n1)×3 , 50=n1 , 152=1+3n

n=51     A1     N2

[3 marks]

b.

Examiners report

Candidates probably had the most success with this question with many good solutions which were written with the working clearly shown. Many used the alternate approach of un=3n1 .

a.

Candidates probably had the most success with this question with many good solutions which were written with the working clearly shown. Many used the alternate approach of un=3n1 .

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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