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Date May 2013 Marks available 2 Reference code 13M.2.hl.TZ2.5
Level HL only Paper 2 Time zone TZ2
Command term Find Question number 5 Adapted from N/A

Question

The arithmetic sequence {un:nZ+} has first term u1=1.6 and common difference d = 1.5. The geometric sequence {vn:nZ+} has first term v1=3 and common ratio r = 1.2.

Find an expression for unvn in terms of n.

[2]
a.

Determine the set of values of n for which un>vn.

[3]
b.

Determine the greatest value of unvn. Give your answer correct to four significant figures.

[1]
c.

Markscheme

unvn=1.6+(n1)×1.53×1.2n1 (=1.5n+0.13×1.2n1)     A1A1

[2 marks]

a.

attempting to solve un>vn numerically or graphically.     (M1)

n=2.621,9.695     (A1)

So 3n9     A1

[3 marks]

b.

The greatest value of unvn is 1.642.     A1

Note: Do not accept 1.64.

 

[1 mark]

c.

Examiners report

In part (a), most candidates were able to express un and vn correctly and hence obtain a correct expression for unvn. Some candidates made careless algebraic errors when unnecessarily simplifying un while other candidates incorrectly stated vn as 3(1.2)n.

a.

In parts (b) and (c), most candidates treated n as a continuous variable rather than as a discrete variable. Candidates should be aware that a GDC’s table feature can be extremely useful when attempting such question types.

b.

In parts (b) and (c), most candidates treated n as a continuous variable rather than as a discrete variable. Candidates should be aware that a GDC’s table feature can be extremely useful when attempting such question types. In part (c), a number of candidates attempted to find the maximum value of n rather than attempting to find the maximum value of unvn.

c.

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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