| 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, \(r\).
Hence, show that \(m\) satisfies the equation \({m^2} + 3m - 40 = 0\).
Find the two possible values of \(m\).
Find the possible values of \(r\).
The sequence has a finite sum.
State which value of \(r\) leads to this sum and justify your answer.
The sequence has a finite sum.
Calculate the sum of the sequence.
Markscheme
correct expression for \(r\) A1 N1
eg \(r = \frac{6}{{m - 1}},{\text{ }}\frac{{m + 4}}{6}\)
[2 marks]
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]
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]
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]
\(r = - \frac{2}{3}\) (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]
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]
