| Date | November 2014 | Marks available | 5 | Reference code | 14N.2.sl.TZ0.5 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
In a game, n small pumpkins are placed 1 metre apart in a straight line. Players start 3 metres before the first pumpkin.
Each player collects a single pumpkin by picking it up and bringing it back to the start. The nearest pumpkin is collected first. The player then collects the next nearest pumpkin and the game continues in this way until the signal is given for the end.
Sirma runs to get each pumpkin and brings it back to the start.
Write down the distance, \({a_1}\), in metres that she has to run in order to collect the first pumpkin.
The distances she runs to collect each pumpkin form a sequence \({a_1},{\text{ }}{a_2},{\text{ }}{a_3}, \ldots \) .
(i) Find \({a_2}\).
(ii) Find \({a_3}\).
Write down the common difference, \(d\), of the sequence.
The final pumpkin Sirma collected was 24 metres from the start.
(i) Find the total number of pumpkins that Sirma collected.
(ii) Find the total distance that Sirma ran to collect these pumpkins.
Peter also plays the game. When the signal is given for the end of the game he has run 940 metres.
Calculate the total number of pumpkins that Peter collected.
Peter also plays the game. When the signal is given for the end of the game he has run 940 metres.
Calculate Peter’s distance from the start when the signal is given.
Markscheme
\(6{\text{ (m)}}\) (A1)(G1)
(i) \(8\) (A1)(ft)
(ii) \(10\) (A1)(ft)(G2)
Note: Follow through from part (a).
\(2{\text{ (m)}}\) (A1)(ft)
Note: Follow through from parts (a) and (b).
(i) \(2 \times 24 = 6 + 2(n - 1)\;\;\;\)OR\(\;\;\;24 = 3 + (n - 1)\) (M1)
Note: Award (M1) for correct substitution in arithmetic sequence formula.
\(n = 22\) (A1)(ft)(G1)
Note: Follow through from parts (a) and (c).
(ii) \(\frac{{(6 + 48)}}{2} \times 22\) (M1)(A1)(ft)
Note: Award (M1) for substitution in arithmetic series formula, (A1)(ft) for correct substitution.
\( = 594\) (A1)(ft)(G2)
Note: Follow through from parts (a) and (d)(i).
\(\frac{{\left[ {2 \times 6 + 2(n - 1)} \right] \times n}}{2} = 940\) (M1)(A1)(ft)
Notes: Award (M1) for substitution in arithmetic series formula, (A1) for their correct substituted formula equated to \(940\). Follow through from parts (a) and (c).
\({n^2} + 5n - 940 = 0\)
\(n = 28.2611 \ldots \)
\(n = 28\) (A1)(ft)(G2)
\(\frac{{\left[ {2 \times 6 + 2(28 - 1)} \right] \times 28}}{2}\) (M1)
Notes: Award (M1) for substituting their \(28\) into the arithmetic series formula.
\( = 16{\text{ (m)}}\) (A1)(ft)(G2)
