Edexcel C1 2006 June — Question 7 7 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2006
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeFind term or common difference
DifficultyModerate -0.8 This is a straightforward application of standard arithmetic sequence formulas (nth term and sum) with two equations and two unknowns. The context is clear, the setup is routine, and solving simultaneous linear equations is basic algebra. Easier than average for A-level.
Spec1.04h Arithmetic sequences: nth term and sum formulae

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 \(a \mathrm {~km}\) and common difference \(d \mathrm {~km}\). 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 \(d\).

Question 7:
AnswerMarks Guidance
\(a+(n-1)d=k\), \(k=9\) or \(11\); \((u_{11}=)\, a+10d=9\)M1, A1c.a.o Use of \(u_n\) to form a linear equation in \(a\) and \(d\). \(a+nd=9\) is M0A0. For \(a+10d=9\)
\(\frac{n}{2}[2a+(n-1)d]=77\) or \(\frac{(a+l)}{2}\times n=77\), \(l=9\) or \(11\)M1 Use of \(S_n\) to form an equation for \(a\) and \(d\) (LHS) or in \(a\) (RHS). For 1st 2 Ms they must write \(n\) or use \(n=11\)
\((S_{11}=)\,\frac{11}{2}(2a+10d)=77\) or \(\frac{(a+9)}{2}\times11=77\)A1 A correct equation based on \(S_n\)
e.g. \(a+10d=9\) and \(a+5d=7\) or \(a+9=14\)M1 Solving simultaneously or (RHS a linear equation in \(a\)). Must lead to \(a=\ldots\) or \(d=\ldots\) and depends on one previous M
\(a=5\) and \(d=0.4\) or exact equivalentA1, A1 For \(a=5\); for \(d=0.4\) (o.e.). ALT: Uses \(\frac{(a+l)}{2}\times n=77\) to get \(a=5\), gets 2nd and 3rd M1A1 i.e. 4/7. MR: Consistent MR of 11 for 9 leading to \(a=3\), \(d=0.8\) scores M1A0M1A0M1A1ftA1ft
## Question 7:

| $a+(n-1)d=k$, $k=9$ or $11$; $(u_{11}=)\, a+10d=9$ | M1, A1c.a.o | Use of $u_n$ to form a linear equation in $a$ and $d$. $a+nd=9$ is M0A0. For $a+10d=9$ |
| $\frac{n}{2}[2a+(n-1)d]=77$ or $\frac{(a+l)}{2}\times n=77$, $l=9$ or $11$ | M1 | Use of $S_n$ to form an equation for $a$ and $d$ (LHS) or in $a$ (RHS). For 1st 2 Ms they must write $n$ or use $n=11$ |
| $(S_{11}=)\,\frac{11}{2}(2a+10d)=77$ or $\frac{(a+9)}{2}\times11=77$ | A1 | A correct equation based on $S_n$ |
| e.g. $a+10d=9$ and $a+5d=7$ or $a+9=14$ | M1 | Solving simultaneously or (RHS a linear equation in $a$). Must lead to $a=\ldots$ or $d=\ldots$ and depends on one previous M |
| $a=5$ and $d=0.4$ or exact equivalent | A1, A1 | For $a=5$; for $d=0.4$ (o.e.). ALT: Uses $\frac{(a+l)}{2}\times n=77$ to get $a=5$, gets 2nd and 3rd M1A1 i.e. 4/7. MR: Consistent MR of 11 for 9 leading to $a=3$, $d=0.8$ scores M1A0M1A0M1A1ftA1ft |

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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 $a \mathrm {~km}$ and common difference $d \mathrm {~km}$.

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 $d$.\\

\hfill \mbox{\textit{Edexcel C1 2006 Q7 [7]}}