| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Standard +0.3 This is a straightforward geometric series question requiring standard techniques: finding common ratio by division, working backwards to find the first term using logarithm properties (log 16 = 2 log 4), and applying the sum formula. The logarithmic context adds minor complexity but the steps are routine for C2 level. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks |
|---|---|
| (i) \(r = \frac{\log_3 16}{\log_3 4} = \frac{\log_3 4^2}{\log_3 4} = \frac{2\log_3 4}{\log_3 4} = 2\) | M2 A1 |
| Answer | Marks |
|---|---|
| \(a = \frac{\log_3 4}{\log_3 4} = \frac{\log_3 2^2}{2} = \frac{2\log_3 2}{2} = \log_3 2\) | M1 A1 |
| (iii) \(S_6 = \frac{(2^6 - 1)\log_3 2}{2 - 1} = 63\log_3 2\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{\lg 2}{\lg 3}\) | M1 A1 | |
| \(\therefore S_6 = 63 \times \frac{\lg 2}{\lg 3} = 39.7\) (3sf) | A1 | (10) |
(i) $r = \frac{\log_3 16}{\log_3 4} = \frac{\log_3 4^2}{\log_3 4} = \frac{2\log_3 4}{\log_3 4} = 2$ | M2 A1 |
(ii) $ar = \log_3 4$
$a = \frac{\log_3 4}{\log_3 4} = \frac{\log_3 2^2}{2} = \frac{2\log_3 2}{2} = \log_3 2$ | M1 A1 |
(iii) $S_6 = \frac{(2^6 - 1)\log_3 2}{2 - 1} = 63\log_3 2$ | M1 A1 |
let $y = \log_3 2 \therefore 3^y = 2$
$y = \frac{\lg 2}{\lg 3}$ | M1 A1 |
$\therefore S_6 = 63 \times \frac{\lg 2}{\lg 3} = 39.7$ (3sf) | A1 | **(10)** |
---7. The second and third terms of a geometric series are $\log _ { 3 } 4$ and $\log _ { 3 } 16$ respectively.\\
(i) Find the common ratio of the series.\\
(ii) Show that the first term of the series is $\log _ { 3 } 2$.\\
(iii) Find, to 3 significant figures, the sum of the first six terms of the series.\\
\hfill \mbox{\textit{OCR C2 Q7 [10]}}