C1 June 2006 Q7
7. An athlete prepares for a race by completing a practice run on each of 11 consecutive days. On each day after the first day, he runs further than he ran on the previous day. The lengths of his 11 practice runs form an arithmetic sequence with first term \(a\) km and common difference \(d\) km.
He runs 9 km on the 11th day, and he runs a total of 77 km over the 11 day period.
Find the value of \(a\) and the value of \(d\). (7)
| Scheme | Marks |
|---|---|
| \(a + (n - 1)d = k\) \(k = 9\) or 11 | M1 |
| \((u_{11} =)\ a + 10d = 9\) | A1c.a.o. |
| \(\dfrac{n}{2}[2a + (n - 1)d] = 77\) or \(\dfrac{(a + l)}{2}\times n = 77\) \(l = 9\) or 11 | M1 |
| \((S_{11} =)\ \dfrac{11}{2}(2a + 10d) = 77\) or \(\dfrac{(a + 9)}{2}\times 11 = 77\) | A1 |
| e.g. \(\begin{aligned} a + 10d &= 9 \\ a + 5d &= 7 \end{aligned}\) or \(a + 9 = 14\) | M1 |
| \(a = 5\) and \(d = 0.4\) or exact equivalent | A1 A1 |
| (7 marks) |
Notes
1st M1 Use of \(u_n\) to form a linear equation in \(a\) and \(d\). \(a + nd = 9\) is M0A0
1st A1 For \(a + 10d = 9\).
2nd M1 Use of \(S_n\) to form an equation for \(a\) and \(d\) (LHS) or in \(a\) (RHS)
2nd A1 A correct equation based on \(S_n\).
For 1st 2 Ms they must write \(n\) or use \(n = 11\).
3rd M1 Solving (LHS simultaneously) or (RHS a linear equation in \(a\))
Must lead to \(a = \ldots\) or \(d = \ldots\) and depends on one previous M
3rd A1 for \(a = 5\)
4th A1 for \(d = 0.4\) (o.e.)
ALT Uses \(\dfrac{(a + l)}{2}\times n = 77\) to get \(a = 5\), gets second and third M1A1 i.e. 4/7
Then uses \(\dfrac{n}{2}[2a + (n - 1)d] = 77\) to get \(d\), gets 1st M1A1 and 4th A1
MR Consistent MR of 11 for 9 leading to \(a = 3\), \(d = 0.8\) scores M1A0M1A0M1A1ftA1ft