| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find common ratio from terms |
| Difficulty | Moderate -0.8 This is a straightforward geometric series question requiring only direct application of standard formulas. Part (a) uses r = T₄/T₃, part (b) multiplies by r again, and part (c) applies the sum to infinity formula. All steps are routine with no problem-solving insight needed, making it easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(r = \frac{27}{36} = \frac{3}{4}\) | M1 A1 | |
| (b) \(= 27 \times \frac{3}{4} = 20\frac{1}{4}\) | M1 A1 | |
| (c) \(a \times (\frac{3}{4})^2 = 36\) | M1 | |
| \(a = 36 \times \frac{16}{9} = 64\) | A1 | |
| \(S_{\infty} = \frac{64}{1 - \frac{3}{4}} = 256\) | M1 A1 | (8 marks) |
**(a)** $r = \frac{27}{36} = \frac{3}{4}$ | M1 A1 |
**(b)** $= 27 \times \frac{3}{4} = 20\frac{1}{4}$ | M1 A1 |
**(c)** $a \times (\frac{3}{4})^2 = 36$ | M1 |
$a = 36 \times \frac{16}{9} = 64$ | A1 |
$S_{\infty} = \frac{64}{1 - \frac{3}{4}} = 256$ | M1 A1 | (8 marks)5. A geometric series has third term 36 and fourth term 27.
Find
\begin{enumerate}[label=(\alph*)]
\item the common ratio of the series,
\item the fifth term of the series,
\item the sum to infinity of the series.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q5 [8]}}