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Date November 2014 Marks available 5 Reference code 14N.2.sl.TZ0.9
Level SL only Paper 2 Time zone TZ0
Command term Find and Hence or otherwise Question number 9 Adapted from N/A

Question

The first two terms of a geometric sequence \({u_n}\) are \({u_1} = 4\) and \({u_2} = 4.2\).

(i)     Find the common ratio.

(ii)     Hence or otherwise, find \({u_5}\).

[5]
a.

Another sequence \({v_n}\) is defined by \({v_n} = a{n^k}\), where \(a,{\text{ }}k \in \mathbb{R}\), and \(n \in {\mathbb{Z}^ + }\), such that \({v_1} = 0.05\) and \({v_2} = 0.25\).

(i)     Find the value of \(a\).

(ii)     Find the value of \(k\).

[5]
b.

Find the smallest value of \(n\) for which \({v_n} > {u_n}\).

[5]
c.

Markscheme

(i)     valid approach     (M1)

eg\(\;\;\;\)\(r = \frac{{{u_2}}}{{{u_1}}},{\text{ }}\frac{4}{{4.2}}\)

\(r = 1.05\;\;\;{\text{(exact)}}\)     A1     N2

(ii)     attempt to substitute into formula, with their \(r\)     (M1)

eg\(\;\;\;\)\(4 \times {1.05^n},{\text{ }}4 \times 1.05 \times 1.05 \ldots \)

correct substitution     (A1)

eg\(\;\;\;\)\(4 \times {1.05^4},{\text{ }}4 \times 1.05 \times 1.05 \times 1.05 \times 1.05\)

\({u_5} = 4.862025{\text{ (exact), }}4.86{\text{ }}[4.86,{\text{ }}4.87]{\text{ }}\)     A1     N2

[5 marks]

a.

(i)     attempt to substitute \(n = 1\)     (M1)

eg\(\;\;\;\)\(0.05 = a \times {1^k}\)

\(a = 0.05\)     A1     N2

(ii)     correct substitution of \(n = 2\) into \({v_2}\)     A1

eg\(\;\;\;\)\(0.25 = a \times {2^k}\)

correct work     (A1)

eg\(\;\;\;\)finding intersection point, \(k = {\log _2}\left( {\frac{{0.25}}{{0.05}}} \right),{\text{ }}\frac{{\log 5}}{{\log 2}}\)

\(2.32192\)

\(k = {\log _2}5\;\;\;{\text{(exact), }}2.32{\text{ }}[2.32,{\text{ }}2.33]\)     A1     N2

[5 marks]

b.

correct expression for \({u_n}\)     (A1)

eg\(\;\;\;\)\(4 \times {1.05^{n - 1}}\)

EITHER

correct substitution into inequality (accept equation)     (A1)

eg\(\;\;\;\)\(0.05 \times {n^k} > 4 \times {1.05^{n - 1}}\)

valid approach to solve inequality (accept equation)     (M1)

eg\(\;\;\;\)finding point of intersection, \(n = 7.57994{\text{ }}(7.59508{\text{ from 2.32)}}\)

\(n = 8\;\;\;\)(must be an integer)     A1     N2

OR

table of values

when \(n = 7,{\text{ }}{u_7} = 5.3604,{\text{ }}{v_7} = 4.5836\)     A1

when \(n = 8,{\text{ }}{u_8} = 5.6284,{\text{ }}{v_8} = 6.2496\)     A1

\(n = 8\;\;\;\)(must be an integer)     A1     N2

[4 marks]

Total [14 marks]

c.

Examiners report

Most candidates answered part (a) correctly.

a.

A surprising number assumed the second sequence to be geometric as well, and thus part (b) was confusing for many. It was quite common that students did not clearly show which work was relevant to part (i) and which to part (ii), thus often losing marks.

b.

Few students successfully completed part (c) as tried to solve algebraically instead of graphically. Those who used the table of values did not always show two sets of values and consequently lost marks.

c.

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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