| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Form and solve quadratic in parameter |
| Difficulty | Standard +0.3 This is a standard geometric progression problem requiring students to use the property that consecutive terms have a constant ratio, leading to a quadratic equation. The algebraic manipulation is straightforward, and finding the sum to infinity is routine application of a formula. While it requires multiple steps across three parts, each step follows standard procedures without requiring novel insight or particularly complex reasoning. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{5k-6}{3k} = \frac{6k-4}{5k-6} \rightarrow (5k-6)^2 = 3k(6k-4)\) | M1 | OR any valid relationship |
| \(25k^2 - 60k + 36 = 18k^2 - 12k \rightarrow 7k^2 - 48k + 36\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = \frac{6}{7}, 6\) | B1B1 | Allow 0.857(1) for \(\frac{6}{7}\) |
| When \(k = \frac{6}{7}\), \(r = -\frac{2}{3}\) | B1 | Must be exact |
| When \(k = 6\), \(r = \frac{4}{3}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(S_\infty = \frac{a}{1-r}\) with \(r = -\frac{2}{3}\) and \(a = 3 \times \frac{6}{7}\) | M1 | Provided \(0 < |
| \(\frac{18}{7} \div \left(1 + \frac{2}{3}\right) = \frac{54}{35}\) or 1.54 | A1 | FT if 0.857(1) has been used in part (ii) |
## Question 9(i):
$\frac{5k-6}{3k} = \frac{6k-4}{5k-6} \rightarrow (5k-6)^2 = 3k(6k-4)$ | **M1** | OR any valid relationship
$25k^2 - 60k + 36 = 18k^2 - 12k \rightarrow 7k^2 - 48k + 36$ | **A1** | AG
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## Question 9(ii):
$k = \frac{6}{7}, 6$ | **B1B1** | Allow 0.857(1) for $\frac{6}{7}$
When $k = \frac{6}{7}$, $r = -\frac{2}{3}$ | **B1** | Must be exact
When $k = 6$, $r = \frac{4}{3}$ | **B1** |
---
## Question 9(iii):
Use of $S_\infty = \frac{a}{1-r}$ with $r = -\frac{2}{3}$ and $a = 3 \times \frac{6}{7}$ | **M1** | Provided $0 < |their - \frac{2}{3}| < 1$
$\frac{18}{7} \div \left(1 + \frac{2}{3}\right) = \frac{54}{35}$ or 1.54 | **A1** | FT if 0.857(1) has been used in part (ii)
---9 The first, second and third terms of a geometric progression are $3 k , 5 k - 6$ and $6 k - 4$, respectively.\\
(i) Show that $k$ satisfies the equation $7 k ^ { 2 } - 48 k + 36 = 0$.\\
(ii) Find, showing all necessary working, the exact values of the common ratio corresponding to each of the possible values of $k$.\\
(iii) One of these ratios gives a progression which is convergent. Find the sum to infinity.\\
\hfill \mbox{\textit{CAIE P1 2019 Q9 [8]}}