| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Form and solve quadratic in parameter |
| Difficulty | Moderate -0.3 This is a straightforward two-part question testing basic definitions of arithmetic and geometric progressions. Part (i) requires forming and solving a simple quadratic equation (x² - 4x = 12), while part (ii) involves using the sum to infinity formula and finding a common ratio. Both parts are routine applications of standard 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 - 4x = 12\) | M1 | \(4x - x^2 = 12\) scores M1A0 |
| \(x = -2\) or \(6\) | A1 | |
| \(3^\text{rd}\) term \(= (-2)^2 + 12 = 16\) or \(6^2 + 12 = 48\) | A1A1 [4] | SC1 for 16, 48 after \(x = 2, -6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r^2 = \frac{x^2}{4x}\left(= \frac{x}{4}\right)\) soi | M1 | |
| \(\frac{4x}{1 - \frac{x}{4}} = 8\) | M1 | Accept use of unsimplified \(\frac{x^2}{4x}\) or \(\frac{4x}{x^2}\) or \(\frac{4}{x}\) |
| \(x = \frac{4}{3}\) or \(r = \frac{1}{3}\) | A1 | |
| \(3^\text{rd}\) term \(= \frac{16}{27}\) (or 0.593) | A1 [4] | |
| ALT: \(\frac{4x}{1-r} = 8 \rightarrow r = 1 - \frac{1}{2}x\) or \(\frac{4x}{1-r}=8 \rightarrow x = 2(1-r)\) | M1 | |
| \(x^2 = 4x\left(1 - \frac{1}{2}x\right)\)   \(r = \frac{2(1-r)}{4}\) | M1 | |
| \(x = \frac{4}{3}\)   \(r = \frac{1}{3}\) | A1 |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - 4x = 12$ | M1 | $4x - x^2 = 12$ scores M1A0 |
| $x = -2$ or $6$ | A1 | |
| $3^\text{rd}$ term $= (-2)^2 + 12 = 16$ or $6^2 + 12 = 48$ | A1A1 [4] | SC1 for 16, 48 after $x = 2, -6$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r^2 = \frac{x^2}{4x}\left(= \frac{x}{4}\right)$ soi | M1 | |
| $\frac{4x}{1 - \frac{x}{4}} = 8$ | M1 | Accept use of unsimplified $\frac{x^2}{4x}$ or $\frac{4x}{x^2}$ or $\frac{4}{x}$ |
| $x = \frac{4}{3}$ or $r = \frac{1}{3}$ | A1 | |
| $3^\text{rd}$ term $= \frac{16}{27}$ (or 0.593) | A1 [4] | |
| **ALT:** $\frac{4x}{1-r} = 8 \rightarrow r = 1 - \frac{1}{2}x$ **or** $\frac{4x}{1-r}=8 \rightarrow x = 2(1-r)$ | M1 | |
| $x^2 = 4x\left(1 - \frac{1}{2}x\right)$   $r = \frac{2(1-r)}{4}$ | M1 | |
| $x = \frac{4}{3}$   $r = \frac{1}{3}$ | A1 | |
---8 The first term of a progression is $4 x$ and the second term is $x ^ { 2 }$.\\
(i) For the case where the progression is arithmetic with a common difference of 12 , find the possible values of $x$ and the corresponding values of the third term.\\
(ii) For the case where the progression is geometric with a sum to infinity of 8 , find the third term.
\hfill \mbox{\textit{CAIE P1 2015 Q8 [8]}}