C1 January 2011 Q6
6. An arithmetic sequence has first term \(a\) and common difference \(d\). The sum of the first 10 terms of the sequence is 162.
(a) Show that \(10a + 45d = 162\) (2)
Given also that the sixth term of the sequence is 17,
(b) write down a second equation in \(a\) and \(d\), (1)
(c) find the value of \(a\) and the value of \(d\). (4)
| Scheme | Marks |
|---|---|
| \(S_{10} = \dfrac{10}{2}\left[2a + 9d\right]\) or \(S_{10} = a + a + d + a + 2d + a + 3d + a + 4d + a + 5d + a + 6d + a + 7d + a + 8d + a + 9d\) | M1 |
| \(162 = 10a + 45d\) * | A1cso |
| (2) |
Notes
M1: for use of \(S_n\) with \(n = 10\)
(corrected from the printed mark scheme: the listed sum is printed with “\(a + 5da + 6d\)” for “\(a + 5d + a + 6d\)”)
| Scheme | Marks |
|---|---|
| \(\left(u_n = a + (n - 1)d \ \Rightarrow\ \right)17 = a + 5d\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(10\times(b)\) gives \(10a + 50d = 170\) (a) is \(\phantom{\times(b)\text{ gives}}\ 10a + 45d = 162\) | M1 |
| Subtract \(\quad 5d = 8 \quad\) so \(d = \underline{1.6}\) o.e. | A1 |
| Solving for \(a\) \(\quad a = 17 - 5d\) | M1 |
| so \(a = \underline{9}\) | A1 |
| (4) | |
| (7 marks) |
Notes
1st M1: for an attempt to eliminate \(a\) or \(d\) from their two linear equations
2nd M1: for using their value of \(a\) or \(d\) to find the other value.
(These notes are headed (b) in the printed mark scheme.)