| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find common ratio from terms |
| Difficulty | Moderate -0.3 This is a straightforward geometric series question requiring basic logarithm laws (log 16 = log 4² = 2log 4) to find r=2, then working backwards to find the first term. Part (c) involves standard GP sum formula application. While it combines logs with GP, the steps are routine and clearly signposted, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(r = \frac{\log_3 16}{\log_3 4} = \frac{\log_3 4^2}{\log_3 4} = \frac{2\log_3 4}{\log_3 4} = 2\) | M2 A1 | |
| (b) \(ar = \log_3 4\) | ||
| \(a = \frac{\log_3 4}{2} = \frac{\log_3 2^2}{2} = \frac{2\log_3 2}{2} = \log_3 2\) | M1 A1 | |
| (c) \(S_n = \frac{(2^6 - 1)\log_3 2}{2 - 1} = 63\log_3 2\) | M1 A1 | |
| \(= 63 \times \frac{\lg 2}{\lg 3} = 39.7\) | M1 A1 | (9 marks) |
(a) $r = \frac{\log_3 16}{\log_3 4} = \frac{\log_3 4^2}{\log_3 4} = \frac{2\log_3 4}{\log_3 4} = 2$ | M2 A1 |
(b) $ar = \log_3 4$ | |
$a = \frac{\log_3 4}{2} = \frac{\log_3 2^2}{2} = \frac{2\log_3 2}{2} = \log_3 2$ | M1 A1 |
(c) $S_n = \frac{(2^6 - 1)\log_3 2}{2 - 1} = 63\log_3 2$ | M1 A1 |
$= 63 \times \frac{\lg 2}{\lg 3} = 39.7$ | M1 A1 | (9 marks)8. The second and third terms of a geometric series are $\log _ { 3 } 4$ and $\log _ { 3 } 16$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the common ratio of the series.
\item Show that the first term of the series is $\log _ { 3 } 2$.
\item Find, to 3 significant figures, the sum of the first six terms of the series.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q8 [9]}}