| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| 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 -1.2 This is a straightforward arithmetic sequence question requiring only basic formula application: part (a) uses the nth term formula to find N (1500 = 1000 + 20(N-1)), and part (b) applies the sum formula for the arithmetic portion plus simple multiplication for the constant portion. Both parts are routine calculations with no problem-solving insight required, making this easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(1000+(N-1)\times 20=1500 \Rightarrow N=\ldots\) | M1 | Uses a correct term formula with \(a=1000\), \(d=20\) and the 1500 in an attempt to find \(N\). Alternatively calculates \(\frac{1500-1000}{20}+1\). |
| \((N=)26\) | A1 | cao (Allow \(n\) or any other letter for \(N\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(S_{26}=\frac{1}{2}(\text{"26"})[2(1000)+(\text{"26"}-1)\times 20]\) or \(S_{26}=\frac{1}{2}(\text{"26"})[1000+1500]\) | M1 | Correct attempt at AP sum with \(n=\) their \(N\), \(a=1000\), \(d=20\) or \(n=\) their \(N\), \(a=1000\), \(l=1500\) |
| \(=32500\) | A1 | Correct sum (may be implied) |
| constant terms \(=(50-N)\times 1500\) or constant terms \(=(50-(N-1))\times 1500\) | M1 | Attempts \((50-N)\times 1500\) or \((50-(N-1))\times 1500\). So if \(n=26\) was used for the previous M, allow the use of 24 or 25 here. |
| \(S_{50}=\text{"24"}\times 1500 + S_{26}\) | ddM1 | Adds their AP sum to constant terms where 50 terms are being considered. Dependent on both previous M's. |
| \(=68500\) | A1 | cao |
## Question 9(a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $1000+(N-1)\times 20=1500 \Rightarrow N=\ldots$ | M1 | Uses a correct term formula with $a=1000$, $d=20$ and the 1500 in an attempt to find $N$. Alternatively calculates $\frac{1500-1000}{20}+1$. |
| $(N=)26$ | A1 | cao (Allow $n$ or any other letter for $N$) |
---
## Question 9(b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $S_{26}=\frac{1}{2}(\text{"26"})[2(1000)+(\text{"26"}-1)\times 20]$ or $S_{26}=\frac{1}{2}(\text{"26"})[1000+1500]$ | M1 | Correct attempt at AP sum with $n=$ their $N$, $a=1000$, $d=20$ or $n=$ their $N$, $a=1000$, $l=1500$ |
| $=32500$ | A1 | Correct sum (may be implied) |
| constant terms $=(50-N)\times 1500$ or constant terms $=(50-(N-1))\times 1500$ | M1 | Attempts $(50-N)\times 1500$ or $(50-(N-1))\times 1500$. So if $n=26$ was used for the previous M, allow the use of 24 or 25 here. |
| $S_{50}=\text{"24"}\times 1500 + S_{26}$ | ddM1 | Adds their AP sum to constant terms where 50 terms are being considered. Dependent on both previous M's. |
| $=68500$ | A1 | cao |
---9. A car manufacturer currently makes 1000 cars each week.
The manufacturer plans to increase the number of cars it makes each week.
The number of cars made will be increased by 20 each week from 1000 in week 1, to 1020 in week 2, to 1040 in week 3 and so on, until 1500 cars are made in week $N$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $N$.
The car manufacturer then plans to continue to make 1500 cars each week.
\item Find the total number of cars that will be made in the first 50 weeks starting from and including week 1.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q9 [7]}}