Date | May 2011 | Marks available | 3 | Reference code | 11M.1.sl.TZ1.11 |
Level | SL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 11 | Adapted from | N/A |
Question
The seventh term, \({u_7}\) , of a geometric sequence is \(108\). The eighth term, \({u_8}\) , of the sequence is \(36\) .
Write down the common ratio of the sequence.
Find \({u_1}\) .
The sum of the first \(k\) terms in the sequence is \(118 096\) . Find the value of \(k\) .
Markscheme
\(r = \frac{{36}}{{108}}\left( {\frac{1}{3}} \right)\) (A1) (C1)
Note: Accept \(0.333\).
[1 mark]
\({u_1}{\left( {\frac{1}{3}} \right)^7} = 36\) (M1)
Note: Award (M1) for correct substitution in formula for nth term of a GP. Accept equivalent forms.
\({u_1} = 78732\) (A1)(ft) (C2)
Notes: Accept \(78 700\). Follow through from their common ratio found in part (a). If \(0.333\) used from part (a) award (M1)(A1)(ft) for an answer of \(79 285\) or \(79 300\) irrespective of whether working is shown.
[2 marks]
\(118096 = \frac{{78732\left( {1 - {{\left( {\frac{1}{3}} \right)}^k}} \right)}}{{\left( {1 - \frac{1}{3}} \right)}}\) (M1)(M1)
Notes: Award (M1) for correct substitution in the sum of a GP formula, (M1) for equating their sum to \(118 096\). Follow through from parts (a) and (b).
OR
Sketch of the function \(y = 78732\frac{{\left( {1 - {{\left( {\frac{1}{3}} \right)}^k}} \right)}}{{\left( {1 - \frac{1}{3}} \right)}}\) (M1)
Indication of point where \(y = 118 096\) (M1)
OR
\(78 732 + 26 244 + 8748 + 2916 + 972 + 324 + 108 + 36 + 12 + 4 = 118 096\) (M1)(M1)
Note: Award (M1) for a list of at least 8 correct terms, (M1) for the sum of the terms equated to \(118 096\).
\(k =10\) (A1)(ft) (C3)
Notes: Follow through from parts (a) and (b). If k is not an integer, do not award final (A1). Accept alternative methods. If \(0.333\) and \(79 285\) used award (M1)(M1)(A1)(ft) for \(k = 5\). If \(0.333\) and \(79 300\) used award (M1)(M1)(A0).
[3 marks]
Examiners report
In part a, many candidates gave the common ratio as 3.
In part a, many candidates gave the common ratio as 3. While they could set up the equation for part c, relatively few succeeded in solving it. Those who arrived at an answer did not always realize that the answer must be an integer.
In part a, many candidates gave the common ratio as 3. While they could set up the equation for part c, relatively few succeeded in solving it. Those who arrived at an answer did not always realize that the answer must be an integer.