| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Standard +0.3 This is a standard geometric series question requiring systematic application of GP formulas. Part (a) involves solving simultaneous equations from ar and ar^4, part (b) is direct formula application, and part (c) requires algebraic manipulation of S_n and S_∞. While multi-step with 12 marks total, it follows predictable patterns without requiring novel insight—slightly easier than average for a substantial C2 question. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(ar = -48\), \(ar^5 = 6\) | B1 | |
| \(r^4 = \frac{6}{-48} = -\frac{1}{8}\) | M1 | |
| \(r = \sqrt[4]{-\frac{1}{8}} = -\frac{1}{2}\) | M1 A1 | |
| \(a = \frac{-48}{-\frac{1}{2}} = 96\) | A1 | |
| (b) \(= \frac{96}{1-(-\frac{1}{2})} = 64\) | M1 A1 | |
| (c) \(S_n = \frac{90[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})} = 64[1-(-\frac{1}{2})^n]\) | M1 A1 | |
| \(S_{\infty} - S_n = 64 - 64[1-(-\frac{1}{2})^n]\) | M1 | |
| \(= 64(-\frac{1}{2})^n = 2^6 \times (-1)^n \times 2^{-n} = (-1)^n \times 2^{6-n}\) | M1 | |
| Difference is magnitude, \(\therefore 2^{6-n}\) | A1 | (12 marks) |
**(a)** $ar = -48$, $ar^5 = 6$ | B1 |
$r^4 = \frac{6}{-48} = -\frac{1}{8}$ | M1 |
$r = \sqrt[4]{-\frac{1}{8}} = -\frac{1}{2}$ | M1 A1 |
$a = \frac{-48}{-\frac{1}{2}} = 96$ | A1 |
**(b)** $= \frac{96}{1-(-\frac{1}{2})} = 64$ | M1 A1 |
**(c)** $S_n = \frac{90[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})} = 64[1-(-\frac{1}{2})^n]$ | M1 A1 |
$S_{\infty} - S_n = 64 - 64[1-(-\frac{1}{2})^n]$ | M1 |
$= 64(-\frac{1}{2})^n = 2^6 \times (-1)^n \times 2^{-n} = (-1)^n \times 2^{6-n}$ | M1 |
Difference is magnitude, $\therefore 2^{6-n}$ | A1 | (12 marks)
**Total: 75 marks**The second and fifth terms of a geometric series are $-48$ and $6$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the first term and the common ratio of the series. [5]
\item Find the sum to infinity of the series. [2]
\item Show that the difference between the sum of the first $n$ terms of the series and its sum to infinity is given by $2^{6-n}$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [12]}}