| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | January |
| Marks | 10 |
| 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 application requiring only basic formula substitution (nth term and sum formulas) with clear real-world context. Part (a) is verification, part (b) is direct sum calculation, and part (c) requires finding A first then comparing two sums—all routine procedures with no conceptual challenges beyond recognizing the arithmetic sequence structure. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(14000 + 8 \times 1500 = 14000 + 12000 = £26000\) | M1 A1* |
| (b) | \(S_n = \frac{n}{2}(a + l) = \frac{9}{2} \times (14000 + 26000)\) | M1 |
| OR \(S_9 = \frac{n}{2}(2a + (n-1)d) = \frac{9}{2} \times (28000 + 8 \times 1500)\) | Alternatively uses \(S_n = \frac{n}{2}(2a + (n-1)d)\) with \(a = 14000\), \(d = 1500\) and \(n = 8\), 9 or 10. Weaker candidates may list the individual salaries. This is acceptable as long as all terms are included. For example \(14000 + 15500 + 17000 + 18500 + 20000 + 21500 + 23000 + 24500 + 26000\). | |
| \(= £180000\) | A1 | Cao (£) 180000. |
| (c) | Use \(a + (n-1)d\) to find \(A\) | M1 |
| \(A + (10-1) \times 1000 = 26000 \Rightarrow A = 17000\) | A1 | \(A = 17000\). Accept \(A = 17000\) written down for 2 marks as long as no incorrect work seen in its calculation. |
| Use \(S_n = \frac{n}{2}(a + l)\) or \(S_n = \frac{n}{2}(2a + (n-1)d)\) to find \(S\) for Anna | M1 A1 | Use \(S_n = \frac{n}{2}(a+l)\) to find \(S\) for Anna. Follow through on their \(A\), but \(l = 26000\) and \(n = 9\), 10 or 11. Alternatively uses \(S_n = \frac{n}{2}(2a + (n-1)d)\) with their numerical value of \(A\), \(d = 1000\) and \(n = 9\), 10 or 11. Accept a series of terms with their value of \(A\), rising in £1000"s up to a maximum of £26000. Anna earns \(S_{10} = \frac{10}{2}(17000 + 26000)\) OR \(S_{10} = \frac{10}{2}(2 \times 17000 + 9 \times 1000)\) in 10 years. This is an intermediate answer. There is no requirement to state the value £215 000. |
| Shelim earns (b) + 26000 in 10 years. This may be scored at the start of part c. | B1 ft | Shelim earns \(180000 + 26000\) in 10 years. This may be scored at the start of part c. = (£206000). |
| Difference = £9000 | A1 | CAO and CSO Difference = £9000. |
(a) | $14000 + 8 \times 1500 = 14000 + 12000 = £26000$ | M1 A1* | Uses $S = a + (n-1)d$ with $a=14000$, $d=1500$ and $n=8$, 9 or 10 in an attempt to find salary in year 9. Accept a sequence written out only if all terms up to year 9 are included—Allow no errors. csa 26000. It is acceptable to write a sequence for both the 2 marks. FYI the terms are $14000, 15500, 17000, 18500, 20000, 21500, 23000, 24500, 26000$.
(b) | $S_n = \frac{n}{2}(a + l) = \frac{9}{2} \times (14000 + 26000)$ | M1 | Uses $S_n = \frac{n}{2}(a+l)$ with $a = 14000$, $l = 26000$ and $n = 8$, 9 or 10. Do not allow ft"s on incorrect $l$"s.
| OR $S_9 = \frac{n}{2}(2a + (n-1)d) = \frac{9}{2} \times (28000 + 8 \times 1500)$ | | Alternatively uses $S_n = \frac{n}{2}(2a + (n-1)d)$ with $a = 14000$, $d = 1500$ and $n = 8$, 9 or 10. Weaker candidates may list the individual salaries. This is acceptable as long as all terms are included. For example $14000 + 15500 + 17000 + 18500 + 20000 + 21500 + 23000 + 24500 + 26000$.
| $= £180000$ | A1 | Cao (£) 180000.
(c) | Use $a + (n-1)d$ to find $A$ | M1 | Use $l = a + (n-1)d$ to find $A$. It must be a full method with $d = 1000$, $l = 26000a = A$ and $n = 9$, 10 or 11 leading to a value for $A$.
| $A + (10-1) \times 1000 = 26000 \Rightarrow A = 17000$ | A1 | $A = 17000$. Accept $A = 17000$ written down for 2 marks as long as no incorrect work seen in its calculation.
| Use $S_n = \frac{n}{2}(a + l)$ or $S_n = \frac{n}{2}(2a + (n-1)d)$ to find $S$ for Anna | M1 A1 | Use $S_n = \frac{n}{2}(a+l)$ to find $S$ for Anna. Follow through on their $A$, but $l = 26000$ and $n = 9$, 10 or 11. Alternatively uses $S_n = \frac{n}{2}(2a + (n-1)d)$ with their numerical value of $A$, $d = 1000$ and $n = 9$, 10 or 11. Accept a series of terms with their value of $A$, rising in £1000"s up to a maximum of £26000. Anna earns $S_{10} = \frac{10}{2}(17000 + 26000)$ OR $S_{10} = \frac{10}{2}(2 \times 17000 + 9 \times 1000)$ in 10 years. This is an intermediate answer. There is no requirement to state the value £215 000.
| Shelim earns (b) + 26000 in 10 years. This may be scored at the start of part c. | B1 ft | Shelim earns $180000 + 26000$ in 10 years. This may be scored at the start of part c. = (£206000).
| Difference = £9000 | A1 | CAO and CSO Difference = £9000.
---\begin{enumerate}
\item Shelim starts his new job on a salary of $\pounds 14000$. He will receive a rise of $\pounds 1500$ a year for each full year that he works, so that he will have a salary of $\pounds 15500$ in year 2 , a salary of $\pounds 17000$ in year 3 and so on. When Shelim's salary reaches $\pounds 26000$, he will receive no more rises. His salary will remain at $\pounds 26000$.\\
(a) Show that Shelim will have a salary of $\pounds 26000$ in year 9 .\\
(b) Find the total amount that Shelim will earn in his job in the first 9 years.
\end{enumerate}
Anna starts her new job at the same time as Shelim on a salary of $\pounds A$. She receives a rise of $\pounds 1000$ a year for each full year that she works, so that she has a salary of $\pounds ( A + 1000 )$ in year $2 , \pounds ( A + 2000 )$ in year 3 and so on. The maximum salary for her job, which is reached in year 10 , is also $\pounds 26000$.\\
(c) Find the difference in the total amount earned by Shelim and Anna in the first 10 years.\\
\hfill \mbox{\textit{Edexcel C1 2014 Q7 [10]}}