Date | November 2013 | Marks available | 2 | Reference code | 13N.1.sl.TZ0.6 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 6 | Adapted from | N/A |
Question
The first term, \({u_1}\), of an arithmetic sequence is \(145\). The fifth term, \({u_5}\), of the sequence is \(113\).
Find the common difference of the sequence.
The \({n^{{\text{th}}}}\) term, \({u_n}\), of the sequence is \(–7\).
Find the value of \(n\).
The \({n^{{\text{th}}}}\) term, \({u_n}\), of the sequence is \(–7\).
Find \({S_{20}}\), the sum of the first twenty terms of the sequence.
Markscheme
\(145 + (5 - 1)d = 113\) (M1)
Note: Award (M1) for correctly substituted AP formula.
OR
\(\frac{{113 - 145}}{4}\) (M1)
\( = - 8\) (A1) (C2)
[2 marks]
\(145 + (n - 1) \times - 8 = - 7\) (M1)
Note: Award (M1) for their correctly substituted AP formula.
If a list is used award (M1) for their correct values down to \(−7\).
\(n = 20\) (A1)(ft) (C2)
Note: Follow through from their part (a).
[2 marks]
\({S_{20}} = \frac{{20}}{2}\left( {2 \times 145 + (20 - 1) \times - 8} \right)\) (M1)
Note: Award (M1) for their correctly substituted sum of an AP formula.
If a list is used award (M1) for their correct terms up to \(1380\)
\( = 1380\) (A1)(ft)
Note: Follow through from their part (a).
OR
\({S_{20}} = \frac{{20}}{2}\left( {145 + ( - 7)} \right)\) (M1)
Note: Award (M1) for correctly substituted sum of an AP formula.
\( = 1380\) (A1) (C2)
Note: If candidates have listed the terms correctly and given the common difference as \(8\), award (M1)(A0) for part (a), (M1)(A0) for an answer of \(−18\) or \(18\) for part (b) and (M1)(A1)(ft) for an answer of \(4420\) in part (c) with working seen.
[2 marks]
Examiners report
Many candidates gave an answer of 8 rather than –8 but were awarded follow through marks in parts (b) and (c) where working was shown. Some candidates appeared unaware that the common difference in both the AP formula for a term and for a sum is multiplied rather than added or subtracted. Candidates who used a list to answer this question were able to gain full marks.
Many candidates gave an answer of 8 rather than –8 but were awarded follow through marks in parts (b) and (c) where working was shown. Some candidates appeared unaware that the common difference in both the AP formula for a term and for a sum is multiplied rather than added or subtracted. Candidates who used a list to answer this question were able to gain full marks.
Many candidates gave an answer of 8 rather than –8 but were awarded follow through marks in parts (b) and (c) where working was shown. Some candidates appeared unaware that the common difference in both the AP formula for a term and for a sum is multiplied rather than added or subtracted. Candidates who used a list to answer this question were able to gain full marks.