| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Show quadratic equation in n |
| Difficulty | Moderate -0.8 This is a straightforward arithmetic sequence question requiring only direct application of standard formulas (nth term and sum). Part (a) is simple substitution, part (b) involves routine algebraic manipulation to reach a given equation, and part (c) is solving a quadratic—all standard C1 techniques with no problem-solving insight required. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(U_{10} = 500 + (10-1) \times 200\) | M1 | Uses \(a + (n-1)d\) with \(a=500\), \(d=200\) and \(n = 9, 10\) or \(11\) |
| \(= £2300\) | A1 | Correct answer with no working scores full marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{n}{2}\{2 \times 500 + (n-1) \times 200\} = 67200\) | M1A1 | M1: Attempt to use \(S = \frac{n}{2}\{2a + (n-1)d\}\) with \(S_n = 67200\), \(a = 500\), \(d = 200\). A1: Correct equation |
| \(n^2 + 4n - 672 = 0\) | dM1A1 | M1: Attempt to remove brackets and collect terms, dependent on previous M1. A1: Correct three term equation in any form |
| \(n^2 + 4n - 24 \times 28 = 0\) | A1 | Replaces 672 with \(24 \times 28\) with equation as printed (including \(= 0\)) with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((n-24)(n+28) = 0 \Rightarrow n = ..\) or \(n(n+4) = 24 \times 28 \Rightarrow n = ..\) | M1 | Solves the given quadratic in an attempt to find \(n\). May use quadratic formula |
| \(n = 24\) | A1 | States that \(n = 24\), or the number of years is 24 |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $U_{10} = 500 + (10-1) \times 200$ | M1 | Uses $a + (n-1)d$ with $a=500$, $d=200$ and $n = 9, 10$ or $11$ |
| $= £2300$ | A1 | Correct answer with no working scores full marks |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{n}{2}\{2 \times 500 + (n-1) \times 200\} = 67200$ | M1A1 | M1: Attempt to use $S = \frac{n}{2}\{2a + (n-1)d\}$ with $S_n = 67200$, $a = 500$, $d = 200$. A1: Correct equation |
| $n^2 + 4n - 672 = 0$ | dM1A1 | M1: Attempt to remove brackets and collect terms, dependent on previous M1. A1: Correct three term equation in any form |
| $n^2 + 4n - 24 \times 28 = 0$ | A1 | Replaces 672 with $24 \times 28$ with equation as printed (including $= 0$) with no errors |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(n-24)(n+28) = 0 \Rightarrow n = ..$ or $n(n+4) = 24 \times 28 \Rightarrow n = ..$ | M1 | Solves the given quadratic in an attempt to find $n$. May use quadratic formula |
| $n = 24$ | A1 | States that $n = 24$, or the number of years is 24 |
---7. Each year, Abbie pays into a savings scheme. In the first year she pays in $\pounds 500$. Her payments then increase by $\pounds 200$ each year so that she pays $\pounds 700$ in the second year, $\pounds 900$ in the third year and so on.
\begin{enumerate}[label=(\alph*)]
\item Find out how much Abbie pays into the savings scheme in the tenth year.
Abbie pays into the scheme for $n$ years until she has paid in a total of $\pounds 67200$.
\item Show that $n ^ { 2 } + 4 n - 24 \times 28 = 0$
\item Hence find the number of years that Abbie pays into the savings scheme.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q7 [9]}}