| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find first term from conditions |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard formulas for geometric and arithmetic progressions. Part (a) uses sum to infinity formula (one step), part (b)(i) uses two simultaneous equations, part (b)(ii) requires recognizing that the (m+1)th term equals zero, and part (b)(iii) applies the sum formula. All parts are routine applications of memorized formulas with minimal 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{100}{1-r} = 2000\) | M1 | Correct formula and attempt to solve |
| \(r = 19/20\) | A1 | |
| \(ar = 95\) | A1\(\sqrt{}\) | For \(100 \times r\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a + 2d = 90,\ a + 4d = 80\) | B1B1 | |
| \(d = -5,\ a = 100\) | ||
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a + md = 0\) | M1 | Or use correct sum formula |
| \(m = 20\) | A1 | \(m = 20\) with no working scores 2 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{n}{2}[200 + (n-1)(-5)] = 0\) | M1 | |
| \(n = 41\) | A1 | \(n = 41\) with no working scores 2; Do not penalise \(n = 0\) |
| [2] |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{100}{1-r} = 2000$ | M1 | Correct formula and attempt to solve |
| $r = 19/20$ | A1 | |
| $ar = 95$ | A1$\sqrt{}$ | For $100 \times r$ |
| **[3]** | | |
### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a + 2d = 90,\ a + 4d = 80$ | B1B1 | |
| $d = -5,\ a = 100$ | | |
| **[2]** | | |
### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a + md = 0$ | M1 | Or use correct sum formula |
| $m = 20$ | A1 | $m = 20$ with no working scores 2 |
| **[2]** | | |
### Part (b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{n}{2}[200 + (n-1)(-5)] = 0$ | M1 | |
| $n = 41$ | A1 | $n = 41$ with no working scores 2; Do not penalise $n = 0$ |
| **[2]** | | |
---9
\begin{enumerate}[label=(\alph*)]
\item A geometric progression has first term 100 and sum to infinity 2000. Find the second term.
\item An arithmetic progression has third term 90 and fifth term 80 .
\begin{enumerate}[label=(\roman*)]
\item Find the first term and the common difference.
\item Find the value of $m$ given that the sum of the first $m$ terms is equal to the sum of the first ( $m + 1$ ) terms.
\item Find the value of $n$ given that the sum of the first $n$ terms is zero.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2010 Q9 [9]}}