| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Form and solve quadratic in parameter |
| Difficulty | Standard +0.3 This is a straightforward geometric series question requiring standard techniques: using the common ratio property to form an equation, verifying a root, factorizing a cubic, and applying the GP sum formula. While it has multiple parts (12 marks total), each step follows routine procedures with no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks |
|---|---|
| \(r = \frac{x+6}{x-2} = \frac{x^2}{x+6}\) | M1 |
| \((x+6)^2 = x^2(x-2)\) | M1 |
| \(x^2 + 12x + 36 = x^3 - 2x^2\), \(x^3 - 3x^2 - 12x - 36 = 0\) | A1 |
| Answer | Marks |
|---|---|
| when \(x = 6\), LHS: \(216 - 108 - 72 - 36 = 0\) \(\therefore x = 6\) is a solution | B1 |
| Answer | Marks |
|---|---|
| \[6x - 36\] | M1 A1 |
| Answer | Marks |
|---|---|
| \(\therefore\) no other solutions | M1 A1 |
| Answer | Marks |
|---|---|
| \(r = \frac{6+6}{6-2} = 3\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 6 - 2 = 4\) | M1 A1 | |
| \(S_8 = \frac{4(3^8-1)}{3-1} = 13120\) | M1 A1 | (12) |
## Part (i)
$r = \frac{x+6}{x-2} = \frac{x^2}{x+6}$ | M1 |
$(x+6)^2 = x^2(x-2)$ | M1 |
$x^2 + 12x + 36 = x^3 - 2x^2$, $x^3 - 3x^2 - 12x - 36 = 0$ | A1 |
## Part (ii)
when $x = 6$, LHS: $216 - 108 - 72 - 36 = 0$ $\therefore x = 6$ is a solution | B1 |
$$x - 6 \mathop{)} \overline{x^3 + 3x + 6}_\text{} x^2 - 3x^2 - 12x - 36$$
$$x^3 - 6x^2$$
$$3x^2 - 12x$$
$$3x^2 - 18x$$
$$6x - 36$$
$$6x - 36$$ | M1 A1 |
$\therefore (x-6)(x^2 + 3x + 6) = 0$
$x = 6$ or $x^2 + 3x + 6 = 0$
$b^2 - 4ac = 3^2 - (4 \times 1 \times 6) = -15$
$b^2 - 4ac < 0$ $\therefore$ no real solutions to quadratic
$\therefore$ no other solutions | M1 A1 |
## Part (iii)
$r = \frac{6+6}{6-2} = 3$ | B1 |
## Part (iv)
$a = 6 - 2 = 4$ | M1 A1 |
$S_8 = \frac{4(3^8-1)}{3-1} = 13120$ | M1 A1 | (12)
---The first three terms of a geometric series are $(x - 2)$, $(x + 6)$ and $x^2$ respectively.
\begin{enumerate}[label=(\roman*)]
\item Show that $x$ must be a solution of the equation
$$x^3 - 3x^2 - 12x - 36 = 0. \quad \text{(I)}$$ [3]
\item Verify that $x = 6$ is a solution of equation (I) and show that there are no other real solutions. [6]
\end{enumerate}
Using $x = 6$,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item find the common ratio of the series, [1]
\item find the sum of the first eight terms of the series. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C2 Q8 [12]}}