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Date November 2013 Marks available 2 Reference code 13N.1.sl.TZ0.9
Level SL only Paper 1 Time zone TZ0
Command term Hence and Show that Question number 9 Adapted from N/A

Question

The first three terms of a infinite geometric sequence are \(m - 1,{\text{ 6, }}m + 4\), where \(m \in \mathbb{Z}\).

Write down an expression for the common ratio, \(r\).

[2]
a(i).

Hence, show that \(m\) satisfies the equation \({m^2} + 3m - 40 = 0\).

[2]
a(ii).

Find the two possible values of \(m\).

[3]
b(i).

Find the possible values of \(r\).

[3]
b(ii).

The sequence has a finite sum.

State which value of \(r\) leads to this sum and justify your answer.

[3]
c(i).

The sequence has a finite sum.

Calculate the sum of the sequence.

[3]
c(ii).

Markscheme

correct expression for \(r\)     A1     N1

eg   \(r = \frac{6}{{m - 1}},{\text{ }}\frac{{m + 4}}{6}\)

[2 marks]

a(i).

correct equation     A1

eg     \(\frac{6}{{m - 1}} = \frac{{m + 4}}{6},{\text{ }}\frac{6}{{m + 4}} = \frac{{m - 1}}{6}\)

correct working     (A1)

eg     \((m + 4)(m - 1) = 36\)

correct working     A1

eg     \({m^2} - m + 4m - 4 = 36,{\text{ }}{m^2} + 3m - 4 = 36\)

\({m^2} + 3m - 40 = 0\)     AG     N0

[2 marks] 

a(ii).

valid attempt to solve     (M1)

eg     \((m + 8)(m - 5) = 0,{\text{ }}m = \frac{{ - 3 \pm \sqrt {9 + 4 \times 40} }}{2}\)

\(m =  - 8,{\text{ }}m = 5\)     A1A1     N3

[3 marks]

b(i).

attempt to substitute any value of \(m\) to find \(r\)     (M1)

eg     \(\frac{6}{{ - 8 - 1}},{\text{ }}\frac{{5 + 4}}{6}\)

\(r = \frac{3}{2},{\text{ }}r =  - \frac{2}{3}\)     A1A1     N3

[3 marks]

b(ii).

\(r =  - \frac{2}{3}\)   (may be seen in justification)     A1

valid reason     R1     N0

eg     \(\left| r \right| < 1,{\text{ }} - 1 < \frac{{ - 2}}{3} < 1\)

 

Notes: Award R1 for \(\left| r \right| < 1\) only if A1 awarded.

[2 marks]

c(i).

finding the first term of the sequence which has \(\left| r \right| < 1\)     (A1)

eg     \( - 8 - 1,{\text{ }}6 \div \frac{{ - 2}}{3}\)

\({u_1} =  - 9\)   (may be seen in formula)     (A1)

correct substitution of \({u_1}\) and their \(r\) into \(\frac{{{u_1}}}{{1 - r}}\), as long as \(\left| r \right| < 1\)     A1

eg     \({S_\infty } = \frac{{ - 9}}{{1 - \left( { - \frac{2}{3}} \right)}},{\text{ }}\frac{{ - 9}}{{\frac{5}{3}}}\)

\({S_\infty } =  - \frac{{27}}{5}{\text{ }}( =  - 5.4)\)     A1     N3

[4 marks] 

c(ii).

Examiners report

[N/A]
a(i).
[N/A]
a(ii).
[N/A]
b(i).
[N/A]
b(ii).
[N/A]
c(i).
[N/A]
c(ii).

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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