| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | New GP from transformation |
| Difficulty | Standard +0.3 This is a straightforward geometric progression question requiring standard techniques: forming an equation from given terms to find r (part a), applying the sum formula S_n (part b), and recognizing that every third term forms a new GP with modified first term and ratio r³ for sum to infinity (part c). All steps are routine applications of GP formulas with no novel insight required, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute to obtain equation \(9r^4 + 14r^2 - 8 = 0\) | B1 | OE |
| Attempt solution of quadratic equation in \(r^2\) to obtain at least one value of \(r\) or \(r^2\) | M1 | Expect \((9r^2-4)(r^2+2)\) |
| \(r = \frac{2}{3}\) only | A1 | SC B1 answer without working |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(a = 2\) and *their* \(r\) in correct formula and attempt to evaluate | M1 | Expect \(\dfrac{2\!\left(1-\left(\frac{2}{3}\right)^{20}\right)}{\left(1-\frac{2}{3}\right)}\) or \(\dfrac{2\!\left(\left(\frac{2}{3}\right)^{20}-1\right)}{\left(\frac{2}{3}-1\right)}\) |
| \(5.998\) | A1 | AWRT and no other value |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Identify \(a_2 = \frac{4}{3}\) and common ratio as \(\frac{8}{27}\) | B1FT | Following *their* \(r\) provided \( |
| Substitute *their* new \(a\) and \(r\) in correct formula for sum to infinity and evaluate | M1 | \( |
| \(\frac{36}{19}\) | A1 | OE. Accept 1.89 or better from 1.894736… |
| Total | 3 |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute to obtain equation $9r^4 + 14r^2 - 8 = 0$ | B1 | OE |
| Attempt solution of quadratic equation in $r^2$ to obtain at least one value of $r$ or $r^2$ | M1 | Expect $(9r^2-4)(r^2+2)$ |
| $r = \frac{2}{3}$ only | A1 | **SC B1** answer without working |
| **Total** | **3** | |
---
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $a = 2$ and *their* $r$ in correct formula and attempt to evaluate | M1 | Expect $\dfrac{2\!\left(1-\left(\frac{2}{3}\right)^{20}\right)}{\left(1-\frac{2}{3}\right)}$ or $\dfrac{2\!\left(\left(\frac{2}{3}\right)^{20}-1\right)}{\left(\frac{2}{3}-1\right)}$ |
| $5.998$ | A1 | AWRT and no other value |
| **Total** | **2** | |
---
## Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Identify $a_2 = \frac{4}{3}$ and common ratio as $\frac{8}{27}$ | B1FT | Following *their* $r$ provided $|r| < 1$. May be implied in the sum to infinity. Allow $\left(\frac{2}{3}\right)^3$ |
| Substitute *their* new $a$ and $r$ in correct formula for sum to infinity and evaluate | M1 | $|r| < 1$ otherwise M0 |
| $\frac{36}{19}$ | A1 | OE. Accept 1.89 or better from 1.894736… |
| **Total** | **3** | |
---10 The geometric progression $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ has first term 2 and common ratio $r$ where $r > 0$. It is given that $\frac { 9 } { 2 } a _ { 5 } + 7 a _ { 3 } = 8$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $r$.
\item Find the sum of the first 20 terms of the geometric progression. Give your answer correct to 4 significant figures.\\
\includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-14_2725_42_134_2008}
\item Find the sum to infinity of the progression $a _ { 2 } , a _ { 5 } , a _ { 8 } , \ldots$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2024 Q10 [8]}}