| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Moderate -0.5 This is a straightforward geometric progression question requiring standard formula application: finding r from ar² and ar⁵, then finding a, then applying sum to infinity formula. The calculations involve fractional powers and negative terms but follow routine procedures with no conceptual challenges beyond basic GP knowledge. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(ar^2 = -108\), \(ar^5 = 32\) | B1 | |
| \(r^3 = \frac{32}{-108} = \left(-\frac{8}{27}\right)\) | M1 | Eliminating \(a\) |
| \(r = \left(-\frac{2}{3}\right)\) or \(-0.666\) or \(-0.667\) | A1 | \(-\frac{2}{3}\) from little or no working \(\rightarrow\) www |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(a = -243\) | B1√ | ft on their \(r\) \(\left(-108 \text{ or } \frac{32}{r^2}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) \(S_\infty = \frac{-243}{1+\frac{2}{3}} = \frac{-729}{5}\) or \(-145.8\) | M1A1 | Accept \(-146\). For M1 \( |
(i) $ar^2 = -108$, $ar^5 = 32$ | B1 |
$r^3 = \frac{32}{-108} = \left(-\frac{8}{27}\right)$ | M1 | Eliminating $a$
$r = \left(-\frac{2}{3}\right)$ or $-0.666$ or $-0.667$ | A1 | $-\frac{2}{3}$ from little or no working $\rightarrow$ www
[3]
(ii) $a = -243$ | B1√ | ft on their $r$ $\left(-108 \text{ or } \frac{32}{r^2}\right)$
[1]
(iii) $S_\infty = \frac{-243}{1+\frac{2}{3}} = \frac{-729}{5}$ or $-145.8$ | M1A1 | Accept $-146$. For M1 $|r|$ must be $< 1$
[2]4 The third term of a geometric progression is - 108 and the sixth term is 32 . Find\\
(i) the common ratio,\\
(ii) the first term,\\
(iii) the sum to infinity.
\hfill \mbox{\textit{CAIE P1 2013 Q4 [6]}}