| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Standard +0.3 Part (a) involves standard geometric progression calculations: finding the first term using the formula for the nth term, then applying the sum to infinity formula. Part (b) requires recognizing that sector angles sum to 360° and using the arithmetic series sum formula, which is a straightforward application once the setup is identified. Both parts are routine multi-step problems slightly above average difficulty due to the combination of topics and arithmetic involved. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(ar = 24\), \(ar^3 = 13\frac{1}{2}\) | B1 | Both needed |
| Eliminates \(a\) (or \(r\)) \(\rightarrow r = \frac{3}{4}\) | M1 | Method of solution |
| \(\rightarrow a = 32\) | A1 | co |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sum to infinity \(= 32 \div \frac{1}{4} = 128\) | M1 A1\(\checkmark\) | Correct formula used; \(\checkmark\) on value of \(r\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = 3\), \(d = 2\) | B1 | Correct value for \(d\) |
| \(\frac{n}{2}(6 + (n-1)2)\) \((= 360)\) | M1 | Correct \(S_n\) used; no need for 360 here |
| \(\rightarrow 2n^2 + 4n - 720 = 0\) | A1 | Correct quadratic |
| \(\rightarrow n = 18\) | A1 | co |
| [4] |
## Question 8:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $ar = 24$, $ar^3 = 13\frac{1}{2}$ | B1 | Both needed |
| Eliminates $a$ (or $r$) $\rightarrow r = \frac{3}{4}$ | M1 | Method of solution |
| $\rightarrow a = 32$ | A1 | co |
| **[3]** | | |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sum to infinity $= 32 \div \frac{1}{4} = 128$ | M1 A1$\checkmark$ | Correct formula used; $\checkmark$ on value of $r$ |
| **[2]** | | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = 3$, $d = 2$ | B1 | Correct value for $d$ |
| $\frac{n}{2}(6 + (n-1)2)$ $(= 360)$ | M1 | Correct $S_n$ used; no need for 360 here |
| $\rightarrow 2n^2 + 4n - 720 = 0$ | A1 | Correct quadratic |
| $\rightarrow n = 18$ | A1 | co |
| **[4]** | | |
---8
\begin{enumerate}[label=(\alph*)]
\item In a geometric progression, all the terms are positive, the second term is 24 and the fourth term is $13 \frac { 1 } { 2 }$. Find
\begin{enumerate}[label=(\roman*)]
\item the first term,
\item the sum to infinity of the progression.
\end{enumerate}\item A circle is divided into $n$ sectors in such a way that the angles of the sectors are in arithmetic progression. The smallest two angles are $3 ^ { \circ }$ and $5 ^ { \circ }$. Find the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2012 Q8 [9]}}