| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Moderate -0.8 This is a straightforward two-part question on geometric progressions requiring only standard formulas: finding r from a₁ and a₄ using a₄ = a₁r³, then applying S∞ = a₁/(1-r). The arithmetic with mixed numbers adds minor computational work but no conceptual challenge. This is easier than average A-level content as it's pure formula application with no problem-solving or insight required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2 \cdot \frac{1}{4} = \frac{1}{3}r^3\); \(r^3 = \frac{9}{4} \times \frac{3}{16} = \frac{27}{64}\); \(r = \frac{3}{4}\) or 0.75 | M1, A1, A1 | [3] |
| (ii) \(S_{\infty} = \frac{\frac{51}{4}}{1-\frac{3}{4}} = \frac{64}{3}\) (or \(21\frac{1}{3}\) or 21.3) | M1, A1 | cao [2] |
(i) $2 \cdot \frac{1}{4} = \frac{1}{3}r^3$; $r^3 = \frac{9}{4} \times \frac{3}{16} = \frac{27}{64}$; $r = \frac{3}{4}$ or 0.75 | M1, A1, A1 | [3]
(ii) $S_{\infty} = \frac{\frac{51}{4}}{1-\frac{3}{4}} = \frac{64}{3}$ (or $21\frac{1}{3}$ or 21.3) | M1, A1 | cao [2]5 The first term of a geometric progression is $5 \frac { 1 } { 3 }$ and the fourth term is $2 \frac { 1 } { 4 }$. Find\\
(i) the common ratio,\\
(ii) the sum to infinity.
\hfill \mbox{\textit{CAIE P1 2012 Q5 [5]}}