| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| 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 application of standard arithmetic sequence formulas (nth term and sum) with clear scaffolding through parts (a), (b), and (c). The context is simple, the algebra is routine, and part (b) is already given to show, making this easier than average for A-level. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a + 29d = 40.75\) or \(a = 40.75 - 29d\) or \(29d = 40.75 - a\) | M1 A1 | Attempt to use \(a + (n-1)d\) with \(n = 30\); must see \(29d\) not just \((30-1)d\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((S_{30}) = \frac{30}{2}(a+l)\) or \(\frac{30}{2}(a+40.75)\) or \(\frac{30}{2}(2a+(30-1)d)\) or \(15(2a+29d)\) | M1 | Attempt to use \(S_n\) formula with \(n = 30\); must see one of the printed forms |
| \(1005 = 15[a + 40.75]\) | A1 cso | Forming equation with 1005 and \(S_n\) and simplifying |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(67 = a + 40.75\), so \(a = £26.25\) or \(2625p\) or \(26\frac{1}{4}\); NOT \(\frac{105}{4}\) | M1 A1 | Attempt to simplify to \(ka = \ldots\) or \(k = a + m\); for \(a = 26.25\) or \(2625p\) or \(26\frac{1}{4}\) |
| \(29d = 40.75 - 26.25 = 14.5\), so \(d = £0.50\) or \(0.5\) or \(50p\) or \(\frac{1}{2}\) | M1 A1 | Correct attempt at linear equation for \(d\); do not accept other fractions other than \(\frac{1}{2}\); if answer in pence a "p" must be seen |
## Question 9:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a + 29d = 40.75$ or $a = 40.75 - 29d$ or $29d = 40.75 - a$ | M1 A1 | Attempt to use $a + (n-1)d$ with $n = 30$; must see $29d$ not just $(30-1)d$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(S_{30}) = \frac{30}{2}(a+l)$ or $\frac{30}{2}(a+40.75)$ or $\frac{30}{2}(2a+(30-1)d)$ or $15(2a+29d)$ | M1 | Attempt to use $S_n$ formula with $n = 30$; must see one of the printed forms |
| $1005 = 15[a + 40.75]$ | A1 cso | Forming equation with 1005 and $S_n$ and simplifying |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $67 = a + 40.75$, so $a = £26.25$ or $2625p$ or $26\frac{1}{4}$; NOT $\frac{105}{4}$ | M1 A1 | Attempt to simplify to $ka = \ldots$ or $k = a + m$; for $a = 26.25$ or $2625p$ or $26\frac{1}{4}$ |
| $29d = 40.75 - 26.25 = 14.5$, so $d = £0.50$ or $0.5$ or $50p$ or $\frac{1}{2}$ | M1 A1 | Correct attempt at linear equation for $d$; do not accept other fractions other than $\frac{1}{2}$; if answer in pence a "p" must be seen |
---\begin{enumerate}
\item A farmer has a pay scheme to keep fruit pickers working throughout the 30 day season. He pays $\pounds a$ for their first day, $\pounds ( a + d )$ for their second day, $\pounds ( a + 2 d )$ for their third day, and so on, thus increasing the daily payment by $\pounds d$ for each extra day they work.
\end{enumerate}
A picker who works for all 30 days will earn $\pounds 40.75$ on the final day.\\
(a) Use this information to form an equation in $a$ and $d$.
A picker who works for all 30 days will earn a total of $\pounds 1005$\\
(b) Show that $15 ( a + 40.75 ) = 1005$\\
(c) Hence find the value of $a$ and the value of $d$.\\
\hfill \mbox{\textit{Edexcel C1 2010 Q9 [8]}}