| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| 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 application of standard arithmetic sequence formulas (nth term and sum) with clear context. Part (a) uses a_n = a + (n-1)d directly, part (b) applies S_n = n/2(2a + (n-1)d), and part (c) sets up a simple equation. All values are given explicitly, requiring only substitution and basic algebra with no problem-solving insight needed. |
| 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 |
| \(a + 9d = 150 + 9 \times 10 = 240\) | M1 A1 | Using \(a + 9d\) with at least one of \(a = 150\) and \(d = 10\). Being "one off" (e.g. \(a + 10d\)) scores M0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}n\{2a + (n-1)d\} = \frac{20}{2}\{2\times150 + 19\times10\} = 4900\) | M1 A1, A1 | Attempting correct sum formula for \(S_{20}\) with at least one of \(a=150\), \(d=10\). 1st A: any fully correct numerical version |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Kevin: \(\frac{1}{2}n\{2a + (n-1)d\} = \frac{20}{2}\{2A + 19\times30\}\) | B1 | Correct expression in terms of \(A\) for Kevin's total |
| Kevin's total \(= 2 \times\) "4900" | M1 | Equating Kevin's total to twice Jill's total. Kevin's total need not be correct but must be a linear function of \(A\) |
| \(\frac{20}{2}\{2A + 19\times30\} = 2\times\text{"4900"}\) | A1ft | Kevin's total (correct, possibly unsimplified) \(= 2\)(Jill's total), ft Jill's total from (b) |
| \(A = 205\) | A1 |
## Question 7:
**(a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a + 9d = 150 + 9 \times 10 = 240$ | M1 A1 | Using $a + 9d$ with at least one of $a = 150$ and $d = 10$. Being "one off" (e.g. $a + 10d$) scores M0 |
**(b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}n\{2a + (n-1)d\} = \frac{20}{2}\{2\times150 + 19\times10\} = 4900$ | M1 A1, A1 | Attempting correct sum formula for $S_{20}$ with at least one of $a=150$, $d=10$. 1st A: any fully correct numerical version |
**(c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Kevin: $\frac{1}{2}n\{2a + (n-1)d\} = \frac{20}{2}\{2A + 19\times30\}$ | B1 | Correct expression in terms of $A$ for Kevin's total |
| Kevin's total $= 2 \times$ "4900" | M1 | Equating Kevin's total to twice Jill's total. Kevin's total need not be correct but must be a linear function of $A$ |
| $\frac{20}{2}\{2A + 19\times30\} = 2\times\text{"4900"}$ | A1ft | Kevin's total (correct, possibly unsimplified) $= 2$(Jill's total), ft Jill's total from (b) |
| $A = 205$ | A1 | |
---7. Jill gave money to a charity over a 20 -year period, from Year 1 to Year 20 inclusive. She gave $\pounds 150$ in Year $1 , \pounds 160$ in Year 2, $\pounds 170$ in Year 3, and so on, so that the amounts of money she gave each year formed an arithmetic sequence.
\begin{enumerate}[label=(\alph*)]
\item Find the amount of money she gave in Year 10.
\item Calculate the total amount of money she gave over the 20 -year period.
Kevin also gave money to the charity over the same 20 -year period.
He gave $\pounds A$ in Year 1 and the amounts of money he gave each year increased, forming an arithmetic sequence with common difference $\pounds 30$. The total amount of money that Kevin gave over the 20 -year period was twice the total amount of money that Jill gave.
\item Calculate the value of $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2010 Q7 [9]}}