| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Standard +0.3 This question combines arithmetic and geometric progressions but uses standard techniques. Part (a) requires setting up and solving a quadratic equation using the AP property (equal differences), which is routine. Part (b) applies the standard sum to infinity formula S = a/(1-r) after finding the common ratio. The algebraic manipulation is straightforward, and both parts follow textbook methods without requiring novel insight or complex multi-step reasoning. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2(2p-6) = p + \frac{p^2}{6} \Rightarrow \frac{p^2}{6} - 3p + 12[=0]\) OR \((2p-6) - \frac{p^2}{6} = p-(2p-6) \Rightarrow \frac{p^2}{6} - 3p + 12[=0]\) OR \(\frac{1}{6}d^2 + d[=0]\) | \*M1 | Correct method leading to formation of a 3-term quadratic in \(p\) (all terms on one side) or 2-term quadratic in \(d\). OE e.g. \(p^2 - 18p + 72[=0]\), \(\frac{1}{2}p^2 - 9p + 36[=0]\). |
| \(p^2 - 18p + 72[=0] \Rightarrow (p-6)(p-12)[=0]\) or \(\frac{18 \pm \sqrt{(-18)^2 - 4(1)(72)}}{2}\) OR \(d\!\left(\frac{1}{6}d + 1\right)[=0] \Rightarrow d = -6\) | DM1 | Solve a 3-term quadratic in \(p\) by factorisation, formula or completing the square, or solve a 2-term quadratic in \(d\) by factorisation. |
| \(p = 12\) only | A1 | Since \(p = 6\) gives \(d = 0\). If \*M1 DM0 then \(p=12\) only, award SC B1, max 2/3 marks. A0 XP if error in either factor and \(p=12\) only. \(p=12\) only by trial and improvement 3/3. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For GP \(r = \dfrac{2p-6}{\dfrac{p^2}{6}} = \dfrac{18}{24} = \dfrac{3}{4}\) | B1 | OE SOI |
| Sum to infinity \(= \dfrac{24}{1-\dfrac{3}{4}} = 96\) | B1 FT | FT *their* value of \(p\) if used correctly to find \(r\) (B0 if '\(p\)' used) provided \( |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2(2p-6) = p + \frac{p^2}{6} \Rightarrow \frac{p^2}{6} - 3p + 12[=0]$ OR $(2p-6) - \frac{p^2}{6} = p-(2p-6) \Rightarrow \frac{p^2}{6} - 3p + 12[=0]$ OR $\frac{1}{6}d^2 + d[=0]$ | **\*M1** | Correct method leading to formation of a 3-term quadratic in $p$ (all terms on one side) or 2-term quadratic in $d$. OE e.g. $p^2 - 18p + 72[=0]$, $\frac{1}{2}p^2 - 9p + 36[=0]$. |
| $p^2 - 18p + 72[=0] \Rightarrow (p-6)(p-12)[=0]$ or $\frac{18 \pm \sqrt{(-18)^2 - 4(1)(72)}}{2}$ OR $d\!\left(\frac{1}{6}d + 1\right)[=0] \Rightarrow d = -6$ | **DM1** | Solve a 3-term quadratic in $p$ by factorisation, formula or completing the square, or solve a 2-term quadratic in $d$ by factorisation. |
| $p = 12$ only | **A1** | Since $p = 6$ gives $d = 0$. If \*M1 DM0 then $p=12$ only, award **SC B1**, max 2/3 marks. A0 XP if error in either factor and $p=12$ only. $p=12$ only by trial and improvement 3/3. |
| | **3** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For GP $r = \dfrac{2p-6}{\dfrac{p^2}{6}} = \dfrac{18}{24} = \dfrac{3}{4}$ | B1 | OE SOI |
| Sum to infinity $= \dfrac{24}{1-\dfrac{3}{4}} = 96$ | B1 FT | FT *their* value of $p$ if used correctly to find $r$ (B0 if '$p$' used) provided $|r|<1$. e.g. $p=18 \Rightarrow \left[S_\infty =\right] \dfrac{54}{1-\dfrac{5}{9}} = 121.5$ |
---6 The first three terms of an arithmetic progression are $\frac { p ^ { 2 } } { 6 } , 2 p - 6$ and $p$.
\begin{enumerate}[label=(\alph*)]
\item Given that the common difference of the progression is not zero, find the value of $p$.
\item Using this value, find the sum to infinity of the geometric progression with first two terms $\frac { p ^ { 2 } } { 6 }$ and $2 p - 6$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q6 [5]}}