| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Form and solve quadratic in parameter |
| Difficulty | Moderate -0.3 This is a straightforward C2 geometric series question requiring students to use the constant ratio property to form a quadratic equation (given the answer to show), solve it, then apply standard formulas. All steps are routine applications of well-practiced techniques with no novel insight required, making it slightly 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Two of: \(a = k+4\), \(ar = k\), \(ar^2 = 2k-15\) | M1 | Initial step; may be implied by next line |
| \(k^2 = (k+4)(2k-15)\), so \(k^2 = 2k^2 + 8k - 15k - 60\) | M1, A1 | Eliminates \(a\) and \(r\); correct expansion |
| \(k^2 - 7k - 60 = 0\) | A1 | cso; needs \(= 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((k-12)(k+5) = 0 \Rightarrow k = 12\) | M1 A1 | Attempt to solve quadratic; ignore \(k = -5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Common ratio: \(\frac{k}{k+4} = \frac{12}{16} = \frac{3}{4}\) or \(0.75\) | M1 A1 | Complete method to find \(r\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{a}{1-r} = \frac{16}{\frac{1}{4}} = 64\) | M1 A1 | Uses \(\frac{a}{1-r}\); answer is 64 cao |
# Question 9:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Two of: $a = k+4$, $ar = k$, $ar^2 = 2k-15$ | M1 | Initial step; may be implied by next line |
| $k^2 = (k+4)(2k-15)$, so $k^2 = 2k^2 + 8k - 15k - 60$ | M1, A1 | Eliminates $a$ and $r$; correct expansion |
| $k^2 - 7k - 60 = 0$ | A1 | cso; needs $= 0$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(k-12)(k+5) = 0 \Rightarrow k = 12$ | M1 A1 | Attempt to solve quadratic; ignore $k = -5$ |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Common ratio: $\frac{k}{k+4} = \frac{12}{16} = \frac{3}{4}$ or $0.75$ | M1 A1 | Complete method to find $r$ |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{a}{1-r} = \frac{16}{\frac{1}{4}} = 64$ | M1 A1 | Uses $\frac{a}{1-r}$; answer is 64 cao |\begin{enumerate}
\item The first three terms of a geometric series are ( $k + 4$ ), $k$ and ( $2 k - 15$ ) respectively, where $k$ is a positive constant.\\
(a) Show that $k ^ { 2 } - 7 k - 60 = 0$.\\
(b) Hence show that $k = 12$.\\
(c) Find the common ratio of this series.\\
(d) Find the sum to infinity of this series.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2009 Q9 [10]}}