| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Real-world AP: find n satisfying a condition |
| Difficulty | Easy -1.3 This is a straightforward arithmetic sequence question requiring only direct application of standard formulas (nth term and sum). Part (a) is simple substitution into a₁ + (n-1)d. Part (b) requires setting up Sₙ equations and solving a linear equation. No problem-solving insight needed, just routine procedural work well below average A-level difficulty. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a + (n-1)d = 600 + 9 \times 120\) | M1 | For \(600 + 9 \times 120\) or \(600 + 8 \times 120\) |
| \(= (\pounds)1680\) | A1 | 1680 with or without the "£" |
| Listing: Lists ten terms starting £600, £720, £840, £960, ... | M1 | Lists ten terms correctly |
| Identifies the \(10^{th}\) term as \((\pounds)1680\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(d = 80\) for Kim | B1 | Identifies or uses \(d = 80\) for Kim |
| \(\frac{N}{2}\{2 \times 600 + (N-1) \times 120\}\) OR \(\frac{N}{2}\{2 \times 130 + (N-1) \times 80\}\) | M1 | Attempts a sum formula for Andy or Kim. Must have \(a=600, d=120\) for Andy or \(a=130, d=80\) for Kim |
| \(\frac{N}{2}\{2 \times 600 + (N-1) \times 120\} = 2 \times \frac{N}{2}\{2 \times 130 + (N-1) \times 80\}\) | A1 | A correct equation in any form |
| \(20N = 360 \Rightarrow N = \ldots\) | dM1 | Proceeds to find value for \(N\). Dependent on first method mark; must be an equation using Andy's and Kim's sum |
| \((N =)18\) | A1 | Ignore \(N/n = 0\); if correct value of \(N\) seen, ignore further reference to years |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a + (n-1)d = 600 + 9 \times 120$ | M1 | For $600 + 9 \times 120$ or $600 + 8 \times 120$ |
| $= (\pounds)1680$ | A1 | 1680 with or without the "£" |
| **Listing:** Lists ten terms starting £600, £720, £840, £960, ... | M1 | Lists ten terms correctly |
| Identifies the $10^{th}$ term as $(\pounds)1680$ | A1 | |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $d = 80$ for Kim | B1 | Identifies or uses $d = 80$ for Kim |
| $\frac{N}{2}\{2 \times 600 + (N-1) \times 120\}$ OR $\frac{N}{2}\{2 \times 130 + (N-1) \times 80\}$ | M1 | Attempts a sum formula for Andy or Kim. Must have $a=600, d=120$ for Andy **or** $a=130, d=80$ for Kim |
| $\frac{N}{2}\{2 \times 600 + (N-1) \times 120\} = 2 \times \frac{N}{2}\{2 \times 130 + (N-1) \times 80\}$ | A1 | A correct equation in any form |
| $20N = 360 \Rightarrow N = \ldots$ | dM1 | Proceeds to find value for $N$. Dependent on first method mark; must be an equation using Andy's and Kim's sum |
| $(N =)18$ | A1 | Ignore $N/n = 0$; if correct value of $N$ seen, ignore further reference to years |\begin{enumerate}
\item Each year, Andy pays into a savings scheme. In year one he pays in $\pounds 600$. His payments increase by $\pounds 120$ each year so that he pays $\pounds 720$ in year two, $\pounds 840$ in year three and so on, so that his payments form an arithmetic sequence.\\
(a) Find out how much Andy pays into the savings scheme in year ten.\\
(2)
\end{enumerate}
Kim starts paying money into a different savings scheme at the same time as Andy. In year one she pays in $\pounds 130$. Her payments increase each year so that she pays $\pounds 210$ in year two, $\pounds 290$ in year three and so on, so that her payments form a different arithmetic sequence.
At the end of year $N$, Andy has paid, in total, twice as much money into his savings scheme as Kim has paid, in total, into her savings scheme.\\
(b) Find the value of $N$.\\
\hfill \mbox{\textit{Edexcel C1 2018 Q4 [7]}}