| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Two related arithmetic progressions |
| Difficulty | Moderate -0.8 This is a straightforward arithmetic sequence problem with clear structure and standard techniques. Part (a) is simple substitution into the nth term formula, part (b) requires equating two expressions and solving a linear equation, and part (c) uses the sum formula for an arithmetic series. All steps are routine C1 content with no problem-solving insight required, making it easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(a + (n-1)d\) with \(a = A\), \(d = d+1\), \(n = 14\) | M1 | |
| \(A + 13(d+1) = A + 13d + 13\) | A1* | (2 marks); given answer — intermediate step \(A+13(d+1)\) and \(A+13d+13\) must both be seen; if brackets missing and formula not stated, e.g. \(A+13d+1 \Rightarrow A+13d+13\) this is M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Calculates time for Yi on Day 14 \(= (A-13)+13(2d-1)\) | M1 | |
| Sets times equal: \(A+13d+13 = (A-13)+13(2d-1) \Rightarrow d = \ldots\) | M1 | |
| \(d = 3\) | A1 cso | (3 marks); needs both M marks; must simplify to 3 (not 39/13) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(\dfrac{n}{2}\{2A + (n-1)D\}\) with \(n=14\) and \(D = d\) or \(d+1\) | M1 | Usually 4 or 3; may also use \(\frac{n}{2}\{A+(A+13D)\}\) |
| Attempts to solve \(\dfrac{14}{2}\{2A + 13\times(d+1)\} = 784 \Rightarrow A = \ldots\) | dM1 | Must use their \(d+1\); allow miscopy of 784 |
| \(A = 30\) | A1 | (3 marks); cao |
# Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $a + (n-1)d$ with $a = A$, $d = d+1$, $n = 14$ | M1 | |
| $A + 13(d+1) = A + 13d + 13$ | A1* | (2 marks); given answer — intermediate step $A+13(d+1)$ and $A+13d+13$ must both be seen; if brackets missing and formula not stated, e.g. $A+13d+1 \Rightarrow A+13d+13$ this is **M0A0** |
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# Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Calculates time for Yi on Day 14 $= (A-13)+13(2d-1)$ | M1 | |
| Sets times equal: $A+13d+13 = (A-13)+13(2d-1) \Rightarrow d = \ldots$ | M1 | |
| $d = 3$ | A1 cso | (3 marks); needs both M marks; must simplify to 3 (not 39/13) |
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# Question 10(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\dfrac{n}{2}\{2A + (n-1)D\}$ with $n=14$ and $D = d$ or $d+1$ | M1 | Usually 4 or 3; may also use $\frac{n}{2}\{A+(A+13D)\}$ |
| Attempts to solve $\dfrac{14}{2}\{2A + 13\times(d+1)\} = 784 \Rightarrow A = \ldots$ | dM1 | Must use their $d+1$; allow miscopy of 784 |
| $A = 30$ | A1 | (3 marks); cao |
**Total: 8 marks**
---\begin{enumerate}
\item Xin has been given a 14 day training schedule by her coach.
\end{enumerate}
Xin will run for $A$ minutes on day 1 , where $A$ is a constant.\\
She will then increase her running time by ( $d + 1$ ) minutes each day, where $d$ is a constant.\\
(a) Show that on day 14 , Xin will run for
$$( A + 13 d + 13 ) \text { minutes. }$$
Yi has also been given a 14 day training schedule by her coach.\\
Yi will run for $( A - 13 )$ minutes on day 1 .\\
She will then increase her running time by ( $2 d - 1$ ) minutes each day.\\
Given that Yi and Xin will run for the same length of time on day 14,\\
(b) find the value of $d$.
Given that Xin runs for a total time of 784 minutes over the 14 days,\\
(c) find the value of $A$.\\
\hfill \mbox{\textit{Edexcel C1 2014 Q10 [8]}}