| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2009 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find common ratio from terms |
| Difficulty | Easy -1.2 This is a straightforward application of standard formulas for arithmetic and geometric progressions. Part (i) requires solving a linear equation using the nth term formula, while part (ii) involves finding a common ratio from ar² = 54 and ar = 96, then calculating the first term. Both are routine textbook exercises requiring only direct formula application with no problem-solving insight needed. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a + d = 96\) and \(a + 3d = 54 \rightarrow d = -21\) \(a = 117\) | B1, M1A1 [3] | For both expressions. Correct method of solution. co (nb no working, \(d\) correct, \(a\) wrong 0/3) |
| (ii) \(ar = 96\) and \(ar^2 = 54 \rightarrow r^2 = \frac{54}{96} \rightarrow r = \frac{3}{4} \rightarrow a = 128\) | B1, M1, A1 [3] | For both expressions. Correct method of solution. co. \(r = \pm\frac{3}{4}\), no penalty. |
**(i)** $a + d = 96$ and $a + 3d = 54 \rightarrow d = -21$ $a = 117$ | B1, M1A1 [3] | For both expressions. Correct method of solution. co (nb no working, $d$ correct, $a$ wrong 0/3)
**(ii)** $ar = 96$ and $ar^2 = 54 \rightarrow r^2 = \frac{54}{96} \rightarrow r = \frac{3}{4} \rightarrow a = 128$ | B1, M1, A1 [3] | For both expressions. Correct method of solution. co. $r = \pm\frac{3}{4}$, no penalty.
**Total: [6]**
---3 A progression has a second term of 96 and a fourth term of 54. Find the first term of the progression in each of the following cases:\\
(i) the progression is arithmetic,\\
(ii) the progression is geometric with a positive common ratio.
\hfill \mbox{\textit{CAIE P1 2009 Q3 [6]}}