| Exam Board | Edexcel |
|---|---|
| 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 standard C2 geometric series question requiring students to use the constant ratio property to form a cubic equation, verify a given root, and apply standard formulas. While it involves multiple steps and algebraic manipulation, all techniques are routine for this level with no novel problem-solving required. The cubic factorization is straightforward once x=6 is verified. |
| 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 |
|---|---|
| \(\begin{array}{c} x^2 - 6x^2 \\ 3x^2 - 12x \\ 3x^2 - 18x \\ 6x - 36 \\ 6x - 36 \end{array}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\therefore\) no other solutions | M1 A1 A1 |
| Answer | Marks |
|---|---|
| \(r = \frac{6+6}{6-2} = 3\) | B1 |
| Answer | Marks |
|---|---|
| \(a = 6 - 2 = 4\) | |
| \(S_n = \frac{4(3^3-1)}{3-1} = 13120\) | M1 A1 |
**(a)**
$r = \frac{x+6}{x-2} = \frac{x^2}{x+6}$ | 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 |
**(b)**
When $x = 6$, LHS $= 216 - 108 - 72 - 36 = 0$ $\therefore x = 6$ is a solution | B1 |
$x - 6 \mid \begin{array}{c} x^2 + 3x + 6 \\ \hline x^3 - 3x^2 - 12x - 36 \end{array}$
$\begin{array}{c} x^2 - 6x^2 \\ 3x^2 - 12x \\ 3x^2 - 18x \\ 6x - 36 \\ 6x - 36 \end{array}$ | 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 A1 |
**(c)**
$r = \frac{6+6}{6-2} = 3$ | B1 |
**(d)**
$a = 6 - 2 = 4$ | |
$S_n = \frac{4(3^3-1)}{3-1} = 13120$ | M1 A1 |9. The first three terms of a geometric series are $( x - 2 ) , ( x + 6 )$ and $x ^ { 2 }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that $x$ must be a solution of the equation
$$x ^ { 3 } - 3 x ^ { 2 } - 12 x - 36 = 0$$
\item Verify that $x = 6$ is a solution of equation (I) and show that there are no other real solutions.
Using $x = 6$,
\item find the common ratio of the series,
\item find the sum of the first eight terms of the series.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [12]}}