| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Mixed arithmetic and geometric |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard techniques: finding GP common ratio from given terms, sum to infinity formula, connecting GP and AP terms to find first term and common difference, and AP sum formula. All steps are routine applications of formulas with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(216r^3 = 64 \rightarrow r = 2/3\) | B1 | Allow decimal to 3sf (AWRT). |
| \(S_\infty = \frac{216}{1 - their\frac{2}{3}} = 648\) cao | M1 A1 | M1 depends on *their* \( |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(216\left(\frac{2}{3}\right) = 144 \rightarrow 144 = a + d\) | B1 FT | SOI, may be implied by \(96 = 144 + 3d\); finding \(a\). Mis-reads not condoned in 9(b). |
| \(216\left(\frac{2}{3}\right)^2 = 96 \rightarrow 96 = a + 4d\) | B1 FT | SOI, may be implied by \(96 = 144 + 3d\) and finding \(a\) |
| Solve simultaneously | \*M1 | No working may be seen |
| \(d = -16\), \(a = 160\) | A1 | Both required |
| \(S_{21} = \frac{21}{2}\{320 + 20(-16)\} = 0\) | DM1 A1 | Or use of \(\frac{21}{2}(a + u_{21})\) |
## Question 9:
### Part 9(a):
$216r^3 = 64 \rightarrow r = 2/3$ | **B1** | Allow decimal to 3sf (AWRT).
$S_\infty = \frac{216}{1 - their\frac{2}{3}} = 648$ cao | **M1 A1** | M1 depends on *their* $|r| < 1$.
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $216\left(\frac{2}{3}\right) = 144 \rightarrow 144 = a + d$ | B1 FT | SOI, may be implied by $96 = 144 + 3d$; finding $a$. Mis-reads not condoned in 9(b). |
| $216\left(\frac{2}{3}\right)^2 = 96 \rightarrow 96 = a + 4d$ | B1 FT | SOI, may be implied by $96 = 144 + 3d$ and finding $a$ |
| Solve simultaneously | \*M1 | No working may be seen |
| $d = -16$, $a = 160$ | A1 | Both required |
| $S_{21} = \frac{21}{2}\{320 + 20(-16)\} = 0$ | DM1 A1 | Or use of $\frac{21}{2}(a + u_{21})$ |
---9 The first term of a geometric progression is 216 and the fourth term is 64.
\begin{enumerate}[label=(\alph*)]
\item Find the sum to infinity of the progression.\\
The second term of the geometric progression is equal to the second term of an arithmetic progression.\\
The third term of the geometric progression is equal to the fifth term of the same arithmetic progression.
\item Find the sum of the first 21 terms of the arithmetic progression.\\
\includegraphics[max width=\textwidth, alt={}, center]{8eb3d21b-dc45-493c-9e5c-3c0535c505e8-14_798_786_269_667}
The diagram shows the circle $x ^ { 2 } + y ^ { 2 } = 2$ and the straight line $y = 2 x - 1$ intersecting at the points $A$ and $B$. The point $D$ on the $x$-axis is such that $A D$ is perpendicular to the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q9 [9]}}