| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Real-world AP: find term or total |
| Difficulty | Moderate -0.8 This is a straightforward arithmetic sequence problem requiring only basic formula application: finding a term using a = 17000 + (k-1)×1500 = 32000, then summing two parts (AP sum formula plus simple multiplication). No problem-solving insight needed, just routine calculation with clearly stated parameters. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(32000 = 17000 + (k-1) \times 1500 \Rightarrow k = \ldots\) | M1 | Use of 32000 with correct formula in attempt to find \(k\). Correct formula may be implied by correct answer. |
| \((k=)\ 11\) | A1 | Cso (Allow \(n=11\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S = \frac{k}{2}(2\times17000+(k-1)\times1500)\) or \(\frac{k}{2}(17000+32000)\) or \(S=\frac{k-1}{2}(2\times17000+(k-2)\times1500)\) or \(\frac{k-1}{2}(17000+30500)\) | M1A1 | M1: Use of correct sum formula with their integer \(n=k\) or \(k-1\) from part (a) where \(3 |
| \(S = \frac{11}{2}(2\times17000+10\times1500)\) or \(\frac{11}{2}(17000+32000)\) \(= 269\,500\) or \(S=\frac{10}{2}(2\times17000+9\times1500)\) or \(\frac{10}{2}(17000+30500) = 237\,500\) | ||
| \(32000 \times \alpha\) where \(\alpha\) is integer, \(3 < \alpha < 18\) | M1 | |
| \(288\,000 + 269\,500 = 557\,500\) or \(320\,000 + 237\,500 = 557\,500\) | ddM1A1 | ddM1: Attempts to add two values, dependent on both previous M's, must be sum of 20 terms i.e. \(\alpha + k = 20\). A1: 557 500 |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $32000 = 17000 + (k-1) \times 1500 \Rightarrow k = \ldots$ | M1 | Use of 32000 with correct formula in attempt to find $k$. Correct formula may be implied by correct answer. |
| $(k=)\ 11$ | A1 | Cso (Allow $n=11$) |
> Note: $32000 = 17000 + 1500k \Rightarrow k=10$ is M0A0. $\frac{32000-17000}{1500}=10 \therefore k=11$ is M1A1.
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S = \frac{k}{2}(2\times17000+(k-1)\times1500)$ or $\frac{k}{2}(17000+32000)$ or $S=\frac{k-1}{2}(2\times17000+(k-2)\times1500)$ or $\frac{k-1}{2}(17000+30500)$ | M1A1 | M1: Use of correct sum formula with their integer $n=k$ or $k-1$ from part (a) where $3<k<20$, $a=17000$, $d=1500$. A1: Any correct un-simplified numerical expression with $n=11$ or $n=10$ |
| $S = \frac{11}{2}(2\times17000+10\times1500)$ or $\frac{11}{2}(17000+32000)$ $= 269\,500$ or $S=\frac{10}{2}(2\times17000+9\times1500)$ or $\frac{10}{2}(17000+30500) = 237\,500$ | | |
| $32000 \times \alpha$ where $\alpha$ is integer, $3 < \alpha < 18$ | M1 | |
| $288\,000 + 269\,500 = 557\,500$ or $320\,000 + 237\,500 = 557\,500$ | ddM1A1 | ddM1: Attempts to add two values, dependent on both previous M's, must be sum of 20 terms i.e. $\alpha + k = 20$. A1: 557 500 |
> Special case: If only $S_{20}$ (£625,000) found in (b), score first M1 only, otherwise apply scheme.Jess started work 20 years ago. In year 1 her annual salary was $\pounds 17000$. Her annual salary increased by $\pounds 1500$ each year, so that her annual salary in year 2 was $\pounds 18500$, in year 3 it was $\pounds 20000$ and so on, forming an arithmetic sequence. This continued until she reached her maximum annual salary of $\pounds 32000$ in year $k$. Her annual salary then remained at $\pounds 32000$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the constant $k$.
\item Calculate the total amount that Jess has earned in the 20 years.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2015 Q9 [7]}}