| Date | November 2016 | Marks available | 3 | Reference code | 16N.1.sl.TZ0.9 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 9 | Adapted from | N/A |
Question
The first two terms of an infinite geometric sequence, in order, are
\(2{\log _2}x,{\text{ }}{\log _2}x\), where \(x > 0\).
The first three terms of an arithmetic sequence, in order, are
\({\log _2}x,{\text{ }}{\log _2}\left( {\frac{x}{2}} \right),{\text{ }}{\log _2}\left( {\frac{x}{4}} \right)\), where \(x > 0\).
Let \({S_{12}}\) be the sum of the first 12 terms of the arithmetic sequence.
Find \(r\).
Show that the sum of the infinite sequence is \(4{\log _2}x\).
Find \(d\), giving your answer as an integer.
Show that \({S_{12}} = 12{\log _2}x - 66\).
Given that \({S_{12}}\) is equal to half the sum of the infinite geometric sequence, find \(x\), giving your answer in the form \({2^p}\), where \(p \in \mathbb{Q}\).
Markscheme
evidence of dividing terms (in any order) (M1)
eg\(\,\,\,\,\,\)\(\frac{{{\mu _2}}}{{{\mu _1}}},{\text{ }}\frac{{2{{\log }_2}x}}{{{{\log }_2}x}}\)
\(r = \frac{1}{2}\) A1 N2
[2 marks]
correct substitution (A1)
eg\(\,\,\,\,\,\)\(\frac{{2{{\log }_2}x}}{{1 - \frac{1}{2}}}\)
correct working A1
eg\(\,\,\,\,\,\)\(\frac{{2{{\log }_2}x}}{{\frac{1}{2}}}\)
\({S_\infty } = 4{\log _2}x\) AG N0
[2 marks]
evidence of subtracting two terms (in any order) (M1)
eg\(\,\,\,\,\,\)\({u_3} - {u_2},{\text{ }}{\log _2}x - {\log _2}\frac{x}{2}\)
correct application of the properties of logs (A1)
eg\(\,\,\,\,\,\)\({\log _2}\left( {\frac{{\frac{x}{2}}}{x}} \right),{\text{ }}{\log _2}\left( {\frac{x}{2} \times \frac{1}{x}} \right),{\text{ }}({\log _2}x - {\log _2}2) - {\log _2}x\)
correct working (A1)
eg\(\,\,\,\,\,\)\({\log _2}\frac{1}{2},{\text{ }} - {\log _2}2\)
\(d = - 1\) A1 N3
[4 marks]
correct substitution into the formula for the sum of an arithmetic sequence (A1)
eg\(\,\,\,\,\,\)\(\frac{{12}}{2}\left( {2{{\log }_2}x + (12 - 1)( - 1)} \right)\)
correct working A1
eg\(\,\,\,\,\,\)\(6(2{\log _2}x - 11),{\text{ }}\frac{{12}}{2}(2{\log _2}x - 11)\)
\(12{\log _2}x - 66\) AG N0
[2 marks]
correct equation (A1)
eg\(\,\,\,\,\,\)\(12{\log _2}x - 66 = 2{\log _2}x\)
correct working (A1)
eg\(\,\,\,\,\,\)\(10{\log _2}x = 66,{\text{ }}{\log _2}x = 6.6,{\text{ }}{2^{66}} = {x^{10}},{\text{ }}{\log _2}\left( {\frac{{{x^{12}}}}{{{x^2}}}} \right) = 66\)
\(x = {2^{6.6}}\) (accept \(p = \frac{{66}}{{10}}\)) A1 N2
[3 marks]
