| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Mixed arithmetic and geometric |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for arithmetic and geometric progressions. Part (i) requires finding the common difference then applying the sum formula, while part (ii) requires finding the common ratio then the nth term formula. Both are routine calculations with no problem-solving insight needed, making this easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a = 12\), \(a + d = 18\), \(\therefore d = 1.5\) | B1 | Correct only |
| \(S_{31} = \frac{25}{2}(24 + 24 \times 1.5) = 750\) | M1, A1 | Use of \(S_n\) formula. Correct only |
| Answer | Marks | Guidance |
|---|---|---|
| 13th term \(= ar^{12} = 12 \times (1.5)^{12} = 40.5\) or \(40.6\) | M1, A1, M1, A1 | Correct method for \(r\) or \(r^4\) (needs \(ar^4\)); Needs \(ar^{12}\) and method for subbing \(r\) (or \(r^4\)); Correct only |
**(i)** $a = 12$, $a + d = 18$, $\therefore d = 1.5$ | B1 | Correct only
$S_{31} = \frac{25}{2}(24 + 24 \times 1.5) = 750$ | M1, A1 | Use of $S_n$ formula. Correct only
**(ii)** $a = 12$, $ar^4 = 18$, $r^4 = 1.5$
13th term $= ar^{12} = 12 \times (1.5)^{12} = 40.5$ or $40.6$ | M1, A1, M1, A1 | Correct method for $r$ or $r^4$ (needs $ar^4$); Needs $ar^{12}$ and method for subbing $r$ (or $r^4$); Correct only4 A progression has a first term of 12 and a fifth term of 18.\\
(i) Find the sum of the first 25 terms if the progression is arithmetic.\\
(ii) Find the 13th term if the progression is geometric.
\hfill \mbox{\textit{CAIE P1 2002 Q4 [7]}}