Date | May 2015 | Marks available | 2 | Reference code | 15M.2.sl.TZ2.3 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The sum of the first \(n\) terms of an arithmetic sequence is given by \({S_n} = 6n + {n^2}\).
Write down the value of
(i) \({S_1}\);
(ii) \({S_2}\).
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
Show that \({u_2} = 9\).
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
Find the common difference of the sequence.
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
Find \({u_{10}}\).
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
Find the lowest value of \(n\) for which \({u_n}\) is greater than \(1000\).
The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).
There is a value of \(n\) for which
\[{u_1} + {u_2} + \ldots + {u_n} = 1512.\]
Find the value of \(n\).
Markscheme
(i) \({S_1} = 7\) (A1)
(ii) \({S_2} = 16\) (A1)
\(({u_2} = ){\text{ }}16 - 7 = 9\) (M1)(AG)
Note: Award (M1) for subtracting 7 from 16. The 9 must be seen.
OR
\(16 - 7 - 7 = 2\)
\(({u_2} = ){\text{ }}7 + (2 - 1)(2) = 9\) (M1)(AG)
Note: Award (M1) for subtracting twice \(7\) from \(16\) and for correct substitution in correct arithmetic sequence formula.
The \(9\) must be seen.
Do not accept: \(9 - 7 = 2,{\text{ }}{u_2} = 7 + (2 - 1)(2) = 9\).
\({u_1} = 7\) (A1)(ft)
\(d = 2{\text{ }}( = 9 - 7)\) (A1)(ft)(G2)
Notes: Follow through from their \({S_1}\) in part (a)(i).
\(7 + 2 \times (10 - 1)\) (M1)
Note: Award (M1) for correct substitution in the correct arithmetic sequence formula. Follow through from their parts (a)(i) and (c).
\( = 25\) (A1)(ft)(G2)
Note: Award (A1)(ft) for their correct tenth term.
\(7 + 2 \times (n - 1) > 1000\) (A1)(ft)(M1)
Note: Award (A1)(ft) for their correct expression for the \({n^{{\text{th}}}}\) term, (M1) for comparing their expression to \(1000\). Accept an equation. Follow through from their parts (a)(i) and (c).
\(n = 498\) (A1)(ft)(G2)
Notes: Answer must be a natural number.
\(6n + {n^2} = 1512\;\;\;\)OR\(\;\;\;\frac{n}{2}\left( {14 + 2(n - 1)} \right) = 1512\;\;\;\)OR
\({S_n} = 1512\;\;\;\)OR\(\;\;\;7 + 9 + \ldots + {u_n} = 1512\) (M1)
Notes: Award (M1) for equating the sum of the first \(n\) terms to \(1512\). Accept a sum of at least the first 7 correct terms.
\(n = 36\) (A1)(G2)
Note: If \(n = 36\) is seen without working, award (G2). Award a maximum of (M1)(A0) if \( - 42\) is also given as a solution.