| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find n given sum condition |
| Difficulty | Moderate -0.3 Part (a) requires solving a simple linear inequality (84 + (n-1)(-3) < 0) for n, which is straightforward. Part (b) involves setting up and solving S_{2k} = S_k using the arithmetic series formula, requiring algebraic manipulation but following a standard method with no novel insight needed. Both parts are slightly easier than average A-level questions due to their routine nature and clear structure. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(84 - 3(n-1) = 0\) | M1 | OE, SOI. Allow either \(= 0\) or \(< 0\) (to \(-3\)). |
| Smallest \(n\) is 30 | A1 | SC B2 for answer only \(n = 30\) WWW. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{2k}{2}\right)[168 + (2k-1)(-3)] = \left(\frac{k}{2}\right)[168 + (k-1)(-3)]\) | M1 A1 | M1 for forming an equation using correct formula. A1 for at least one side correct. |
| \(k = 19\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $84 - 3(n-1) = 0$ | M1 | OE, SOI. Allow either $= 0$ or $< 0$ (to $-3$). |
| Smallest $n$ is 30 | A1 | **SC B2** for answer only $n = 30$ WWW. |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{2k}{2}\right)[168 + (2k-1)(-3)] = \left(\frac{k}{2}\right)[168 + (k-1)(-3)]$ | M1 A1 | M1 for forming an equation using correct formula. A1 for at least one side correct. |
| $k = 19$ | A1 | |
---4 The first term of an arithmetic progression is 84 and the common difference is - 3 .
\begin{enumerate}[label=(\alph*)]
\item Find the smallest value of $n$ for which the $n$th term is negative.\\
It is given that the sum of the first $2 k$ terms of this progression is equal to the sum of the first $k$ terms.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q4 [5]}}