| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Shared terms between AP and GP |
| Difficulty | Standard +0.3 This question requires setting up equations linking AP and GP terms using standard formulas (a+8d=8r, a+20d=8r²), then solving a quadratic. It's slightly above average difficulty due to the algebraic manipulation and coordination between two sequences, but follows a standard problem type with clear structure and routine techniques. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (i) GP 8 — \(8r\) — \(8r^2\) AP 8 — \(8+8d\) — \(8+20d\) \(8r = 8+8d\) and \(8r^2 = 8+20d\) Eliminates \(d\) → \(2r^2−5r+3=0\) → \(r = 1.5\) (or 1) | B1 B1 M1 A1 [4] | B1 for each equation. Correct elimination. co (no penalty for including \(r = 1\)) |
| (ii) 4th term of GP = \(ar^3 = 8 \times 27/8 = 27\) If \(r = 1.5, d = 0.5\) 4th term of AP = \(a+3d = 9\frac{1}{2}\) | B1✓ M1A1 [3] | co needs \(a+3d\) and correct method for \(d\) |
**(i)** GP 8 — $8r$ — $8r^2$ AP 8 — $8+8d$ — $8+20d$ $8r = 8+8d$ and $8r^2 = 8+20d$ Eliminates $d$ → $2r^2−5r+3=0$ → $r = 1.5$ (or 1) | B1 B1 M1 A1 [4] | B1 for each equation. Correct elimination. co (no penalty for including $r = 1$)
**(ii)** 4th term of GP = $ar^3 = 8 \times 27/8 = 27$ If $r = 1.5, d = 0.5$ 4th term of AP = $a+3d = 9\frac{1}{2}$ | B1✓ M1A1 [3] | co needs $a+3d$ and correct method for $d$6 The 1st, 2nd and 3rd terms of a geometric progression are the 1st, 9th and 21st terms respectively of an arithmetic progression. The 1st term of each progression is 8 and the common ratio of the geometric progression is $r$, where $r \neq 1$. Find\\
(i) the value of $r$,\\
(ii) the 4th term of each progression.
\hfill \mbox{\textit{CAIE P1 2014 Q6 [7]}}