| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Show quadratic equation in n |
| Difficulty | Moderate -0.8 This is a straightforward arithmetic sequence question requiring standard formula application. Parts (a)-(b) use direct nth term and sum formulas, part (c) involves algebraic manipulation to reach a given quadratic (with the answer provided), and parts (d)-(e) are routine quadratic solving and interpretation. All steps are textbook exercises with no problem-solving insight required. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t_{40} = 100 + (40-1) \times 5\) | M1 | Uses \(a + (n-1)d\) with \(a = 100\), \(d = 5\), \(n = 40\). May be implied by correct expression e.g. \(100 + 39 \times 5\) |
| \(= \text{£}295\) | A1 | Cao. Correct answer with no working scores both marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{60} = \frac{1}{2}(60)(2 \times 100 + (60-1) \times 5)\) or \(l = 100 + (60-1) \times 5 = 395\), \(S_{60} = \frac{1}{2}(60)(100+395)\) | M1 | Uses correct sum formula with \(a = 100\), \(d = 5\), \(n = 60\) or \(n = 40\). If using \(\frac{1}{2}n(a+l)\) with \(n = 40\): \(\frac{1}{2}(40)(100+295)\) using result from (a) also scores M1 |
| Correct numerical expression with \(n = 60\) | A1 | If missing brackets mark withheld unless correct expression implied by answer |
| \(= \text{£}14\,850\) | A1 | Cao. Correct answer with no working scores 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}n(2 \times 600 + (n-1) \times -10) = 18200\) | M1 | Attempts correct sum formula with \(a = 600\), \(d = -10\), sets \(= 18\,200\). Condone poor use of brackets |
| Correct equation | A1 | May be implied by subsequent work |
| \(600n - 5n^2 + 5n = 18200\) → \(5n^2 - 605n + 18200 = 0\) → \(n^2 - 121n + 3640 = 0\) | A1* | Obtains printed answer with at least one intermediate line, no errors. Allow other variables for \(n\) but final answer must include "\(= 0\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((n-56)(n-65) = 0 \Rightarrow n = 56, 65\) | M1 | Attempts to solve given quadratic. Must reach at least one value for \(n\) (allow use of \(x\)) |
| \(n = 56, 65\) | A1 | Correct values, ignore labelling e.g. \(x = \ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n = 65\) because e.g. the money has already been saved after 56 months | B1 | States \(n = 65\) with suitable reason. No contradictory statements; any calculations must be correct |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t_{40} = 100 + (40-1) \times 5$ | M1 | Uses $a + (n-1)d$ with $a = 100$, $d = 5$, $n = 40$. May be implied by correct expression e.g. $100 + 39 \times 5$ |
| $= \text{£}295$ | A1 | Cao. Correct answer with no working scores both marks. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{60} = \frac{1}{2}(60)(2 \times 100 + (60-1) \times 5)$ or $l = 100 + (60-1) \times 5 = 395$, $S_{60} = \frac{1}{2}(60)(100+395)$ | M1 | Uses correct sum formula with $a = 100$, $d = 5$, $n = 60$ **or** $n = 40$. If using $\frac{1}{2}n(a+l)$ with $n = 40$: $\frac{1}{2}(40)(100+295)$ using result from (a) also scores M1 |
| Correct numerical expression with $n = 60$ | A1 | If missing brackets mark withheld unless correct expression implied by answer |
| $= \text{£}14\,850$ | A1 | Cao. Correct answer with no working scores 3 marks |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}n(2 \times 600 + (n-1) \times -10) = 18200$ | M1 | Attempts correct sum formula with $a = 600$, $d = -10$, sets $= 18\,200$. Condone poor use of brackets |
| Correct equation | A1 | May be implied by subsequent work |
| $600n - 5n^2 + 5n = 18200$ → $5n^2 - 605n + 18200 = 0$ → $n^2 - 121n + 3640 = 0$ | A1* | Obtains printed answer with at least one intermediate line, no errors. Allow other variables for $n$ but final answer must include "$= 0$" |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(n-56)(n-65) = 0 \Rightarrow n = 56, 65$ | M1 | Attempts to solve given quadratic. Must reach at least one value for $n$ (allow use of $x$) |
| $n = 56, 65$ | A1 | Correct values, ignore labelling e.g. $x = \ldots$ |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n = 65$ because e.g. the money has already been saved after 56 months | B1 | States $n = 65$ with suitable reason. No contradictory statements; any calculations must be correct |
---5. Ben is saving for the deposit for a house over a period of 60 months.
Ben saves $\pounds 100$ in the first month and in each subsequent month, he saves $\pounds 5$ more than the previous month, so that he saves $\pounds 105$ in the second month, $\pounds 110$ in the third month, and so on, forming an arithmetic sequence.
\begin{enumerate}[label=(\alph*)]
\item Find the amount Ben saves in the 40th month.
\item Find the total amount Ben saves over the 60 -month period.
Lina is also saving for a deposit for a house.\\
Lina saves $\pounds 600$ in the first month and in each subsequent month, she saves $\pounds 10$ less than the previous month, so that she saves $\pounds 590$ in the second month, $\pounds 580$ in the third month, and so on, forming an arithmetic sequence.
Given that, after $n$ months, Lina will have saved exactly $\pounds 18200$ for her deposit,
\item form an equation in $n$ and show that it can be written as
$$n ^ { 2 } - 121 n + 3640 = 0$$
\item Solve the equation in part (c).
\item State, with a reason, which of the solutions to the equation in part (c) is not a sensible value for $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2020 Q5 [11]}}