| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Show quadratic equation in n |
| Difficulty | Moderate -0.3 This is a straightforward arithmetic sequence problem requiring standard formulas. Part (a) uses the nth term formula (trivial verification), part (b) applies the sum formula to set up a quadratic equation (routine algebraic manipulation), and part (c) solves the quadratic. All steps are mechanical applications of well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((u_{12} =) 400 + 11 \times -10 = 290\) | B1* | Correct working to obtain 290; must be a correct calculation |
| Alt 1: \(400 + (n-1) \times -10 = 290 \Rightarrow n = 12\) | B1* | Correct working using 290 to obtain \(n=12\); must have at least one intermediate line |
| Alt 2: \(290 = 400 + (12-1)d \Rightarrow d = -10\) | B1* | Correct working using 290 and 400 to obtain \(d = -10\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8100 = \frac{1}{2}N(2 \times 400 + (N-1) \times -10)\) | M1 | Uses correct sum formula in terms of \(N\) with \(a=400\), \(d=-10\) and sets \(= 8100\) |
| \(\Rightarrow N^2 - 81N + 1620 = 0\) | A1* | Fully correct proof with sufficient working; all brackets removed to obtain unsimplified expanded quadratic before given answer including "\(= 0\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N^2 - 81N + 1620 = 0 \Rightarrow (N-45)(N-36) = 0 \Rightarrow N = 45, 36\) | M1 | Solves given quadratic by any correct method to obtain at least one value of \(N\) |
| \((N =)\ 36\) | A1 | Realises smaller value required; must be clear this value has been selected; \(N=36\) with no working scores M1A1 |
## Question 2:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(u_{12} =) 400 + 11 \times -10 = 290$ | B1* | Correct working to obtain 290; must be a correct calculation |
| **Alt 1:** $400 + (n-1) \times -10 = 290 \Rightarrow n = 12$ | B1* | Correct working using 290 to obtain $n=12$; must have at least one intermediate line |
| **Alt 2:** $290 = 400 + (12-1)d \Rightarrow d = -10$ | B1* | Correct working using 290 and 400 to obtain $d = -10$ |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8100 = \frac{1}{2}N(2 \times 400 + (N-1) \times -10)$ | M1 | Uses correct sum formula in terms of $N$ with $a=400$, $d=-10$ and sets $= 8100$ |
| $\Rightarrow N^2 - 81N + 1620 = 0$ | A1* | Fully correct proof with sufficient working; all brackets removed to obtain unsimplified expanded quadratic before given answer including "$= 0$" |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N^2 - 81N + 1620 = 0 \Rightarrow (N-45)(N-36) = 0 \Rightarrow N = 45, 36$ | M1 | Solves given quadratic by any correct method to obtain at least one value of $N$ |
| $(N =)\ 36$ | A1 | Realises smaller value required; must be clear this value has been selected; $N=36$ with no working scores M1A1 |
---\begin{enumerate}
\item Jamie takes out an interest-free loan of $\pounds 8100$
\end{enumerate}
Jamie makes a payment every month to pay back the loan.\\
Jamie repays $\pounds 400$ in month $1 , \pounds 390$ in month $2 , \pounds 380$ in month 3 , and so on, so that the amounts repaid each month form an arithmetic sequence.\\
(a) Show that Jamie repays $\pounds 290$ in month 12
After Jamie's $N$ th payment, the loan is completely paid back.\\
(b) Show that $N ^ { 2 } - 81 N + 1620 = 0$\\
(c) Hence find the value of $N$.
\hfill \mbox{\textit{Edexcel Paper 2 2024 Q2 [5]}}