| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find term or common difference |
| Difficulty | Moderate -0.8 This is a straightforward arithmetic sequence problem requiring only direct application of standard formulas. Students identify n=10 and n=40, set up two equations (a+9d=2400, a+39d=600), solve simultaneously for d and a, then apply the sum formula S_40. All steps are routine with no conceptual challenges or problem-solving insight required. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a+9d=2400\) and \(a+39d=600\) | M1 | For attempt to use 2400 and 600 in \(a+(n-1)d\). Need \(a+pd=2400\) and \(a+qd=600\) where \(p=8\) or \(9\) and \(q=38\) or \(39\) |
| \(d=\dfrac{-1800}{30},\ d=-60\) (accept \(\pm60\)) | M1 A1 | 2nd M1 for attempt to solve their 2 linear equations; A1 for \(d=\pm60\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a - 540 = 2400,\ a = 2940\) | M1 A1 | M1 for use of their \(d\) in correct linear equation to find \(a\); A1 their \(a\) compatible with \(d\), so \(d=60\) must give \(a=600\), \(d=-60\) gives \(a=2940\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Total} = \dfrac{1}{2}n\{2a+(n-1)d\} = \dfrac{1}{2}\times40\times(5880+39\times-60)\) | M1 A1ft | M1 for correct \(S_n\) formula with \(n=40\) and at least one of \(a\), \(d\) or \(l\) correct or correct ft; A1ft for correct \(S_{40}\) formula with both \(a,d\) or \(a,l\) correct or correct ft. ALT: \(\frac{1}{2}n\{a+l\}=\frac{1}{2}\times40\times(2940+600)\) |
| \(= 70\,800\) | A1cao | For 70800 only |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a+9d=2400$ and $a+39d=600$ | M1 | For attempt to use 2400 and 600 in $a+(n-1)d$. Need $a+pd=2400$ and $a+qd=600$ where $p=8$ or $9$ and $q=38$ or $39$ |
| $d=\dfrac{-1800}{30},\ d=-60$ (accept $\pm60$) | M1 A1 | 2nd M1 for attempt to solve their 2 linear equations; A1 for $d=\pm60$ |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a - 540 = 2400,\ a = 2940$ | M1 A1 | M1 for use of their $d$ in correct linear equation to find $a$; A1 their $a$ compatible with $d$, so $d=60$ must give $a=600$, $d=-60$ gives $a=2940$ |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Total} = \dfrac{1}{2}n\{2a+(n-1)d\} = \dfrac{1}{2}\times40\times(5880+39\times-60)$ | M1 A1ft | M1 for correct $S_n$ formula with $n=40$ and at least one of $a$, $d$ or $l$ correct or correct ft; A1ft for correct $S_{40}$ formula with both $a,d$ or $a,l$ correct or correct ft. ALT: $\frac{1}{2}n\{a+l\}=\frac{1}{2}\times40\times(2940+600)$ |
| $= 70\,800$ | A1cao | For 70800 only |5. A 40-year building programme for new houses began in Oldtown in the year 1951 (Year 1) and finished in 1990 (Year 40).
The numbers of houses built each year form an arithmetic sequence with first term $a$ and common difference $d$.
Given that 2400 new houses were built in 1960 and 600 new houses were built in 1990, find
\begin{enumerate}[label=(\alph*)]
\item the value of $d$,
\item the value of $a$,
\item the total number of houses built in Oldtown over the 40-year period.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2009 Q5 [8]}}