| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Prove sum formula |
| Difficulty | Moderate -0.8 This is a straightforward application of arithmetic sequence formulas with clear structure and standard algebraic manipulation. Part (a) is a 'show that' requiring basic sum formula application, parts (b) and (c) involve simple equation solving with no conceptual challenges beyond routine C1 arithmetic sequences. |
| 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 |
| \(S_{10} = \frac{10}{2}[2P + 9 \times 2T]\) or \(\frac{10}{2}(P + [P+18T])\) | M1 | For identifying \(a=P\) or \(d=2T\) and attempt at \(S_{10}\); using \(n=10\) and one of \(a\) or \(d\) correct; must see evidence for M mark |
| e.g. \(5[2P+18T] = (£)(10P + 90T)\) | A1cso | For simplifying to given answer; no incorrect working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_{10} = \frac{10}{2}[2(P+1800)+9T] = \{10P+18000+45T\}\) | M1A1 | M1 for attempting \(S_{10}\) for scheme 2 (allow missing brackets); using \(n=10\) and at least one of \(a\) or \(d\) correct; A1 for correct expression |
| \(10P + 90T = 10P + 18000 + 45T\) | M1 | For forming equation using two sums to eliminate \(P\) |
| \(90T = 18000 + 45T\) | ||
| \(T = 400\) (only) | A1 | Answer only (4/4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Scheme 2, Year 10 salary: \([a+(n-1)d =](P+1800)+9T\) | B1ft | For using \(u_{10}\) for scheme 2; can be \(9T\) or follow through their value of \(T\) |
| \(P + 1800 + \text{"3600"} = 29850\) | M1 | For forming equation based on \(u_{10}\) for scheme 2, using 29850 and their value of \(T\) |
| \(P = (£)\ 24450\) | A1 | Answer only (3/3) |
## Question 9:
**(a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_{10} = \frac{10}{2}[2P + 9 \times 2T]$ or $\frac{10}{2}(P + [P+18T])$ | M1 | For identifying $a=P$ or $d=2T$ and attempt at $S_{10}$; using $n=10$ and one of $a$ or $d$ correct; must see evidence for M mark |
| e.g. $5[2P+18T] = (£)(10P + 90T)$ | A1cso | For simplifying to given answer; no incorrect working seen |
**(b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_{10} = \frac{10}{2}[2(P+1800)+9T] = \{10P+18000+45T\}$ | M1A1 | M1 for attempting $S_{10}$ for scheme 2 (allow missing brackets); using $n=10$ and at least one of $a$ or $d$ correct; A1 for correct expression |
| $10P + 90T = 10P + 18000 + 45T$ | M1 | For forming equation using two sums to eliminate $P$ |
| $90T = 18000 + 45T$ | | |
| $T = 400$ (only) | A1 | Answer only (4/4) |
**(c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Scheme 2, Year 10 salary: $[a+(n-1)d =](P+1800)+9T$ | B1ft | For using $u_{10}$ for scheme 2; can be $9T$ or follow through their value of $T$ |
| $P + 1800 + \text{"3600"} = 29850$ | M1 | For forming equation based on $u_{10}$ for scheme 2, using 29850 and their value of $T$ |
| $P = (£)\ 24450$ | A1 | Answer only (3/3) |
---\begin{enumerate}
\item A company offers two salary schemes for a 10 -year period, Year 1 to Year 10 inclusive.
\end{enumerate}
Scheme 1: Salary in Year 1 is $\pounds P$.\\
Salary increases by $\pounds ( 2 T )$ each year, forming an arithmetic sequence.\\
Scheme 2: Salary in Year 1 is $\pounds ( P + 1800 )$.\\
Salary increases by $\pounds T$ each year, forming an arithmetic sequence.\\
(a) Show that the total earned under Salary Scheme 1 for the 10-year period is
$$\pounds ( 10 P + 90 T )$$
For the 10-year period, the total earned is the same for both salary schemes.\\
(b) Find the value of $T$.
For this value of $T$, the salary in Year 10 under Salary Scheme 2 is $\pounds 29850$\\
(c) Find the value of $P$.\\
\hfill \mbox{\textit{Edexcel C1 2012 Q9 [9]}}