| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find first term from conditions |
| Difficulty | Moderate -0.8 This is a straightforward application of standard geometric series formulas (S_n and S_∞) with direct substitution. The algebra is simple, requiring only solving a linear equation after substituting known values. It's easier than average because it's purely procedural with no problem-solving or conceptual challenges. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{a(1-(\frac{1}{3})^1)}{1-\frac{1}{3}} = 200\) | M1 A1 | |
| \(a = 200 \times \frac{27}{40} = 135\) | A1 | |
| (b) \(= \frac{135}{1-\frac{1}{3}} = 202\frac{1}{2}\) | M1 A1 |
**(a)** $\frac{a(1-(\frac{1}{3})^1)}{1-\frac{1}{3}} = 200$ | M1 A1 |
$a = 200 \times \frac{27}{40} = 135$ | A1 |
**(b)** $= \frac{135}{1-\frac{1}{3}} = 202\frac{1}{2}$ | M1 A1 | | (5 marks)A geometric series has common ratio $\frac{1}{3}$.
Given that the sum of the first four terms of the series is 200,
\begin{enumerate}[label=(\alph*)]
\item find the first term of the series, [3]
\item find the sum to infinity of the series. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q2 [5]}}