| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| Marks | 9 |
| 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 standard geometric sequence problem requiring students to use the property that consecutive terms have a constant ratio, leading to a quadratic equation. While it involves multiple parts including sum to infinity, each step follows routine procedures taught in C2 with no novel insight required. Slightly easier than average due to the 'show that' scaffolding in part (a). |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Leave blank | Q16 |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2k-24}{k} = \frac{k}{k+5}\) or \(\frac{k+5}{k} = \frac{k}{2k-24}\) or \((2k-24)(k+5) = k^2\) | M1 | Correct method. I.e. a method that uses the fact that the 3 terms are in geometric progression to establish an equation in \(k\). |
| \((2k-24)(k+5) = 2k^2 - 14k - 120\) | dM1 | Expands \((2k-24)(k+5)\). Must be an attempt at the full expansion but allow the \(k\) terms to be combined. Dependent on the first M. |
| \(2k^2 - 14k - 120 = k^2 \Rightarrow k^2 - 14k - 120 = 0\)* | A1* | Correct solution with no errors including bracketing errors e.g. \(2k - 24(k+5) = \ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((k+6)(k-20) = 0 \Rightarrow k = \ldots\) | M1 | Attempts to solve the given quadratic. See General Guidance. |
| \(k = -6, \ 20\) | A1 | Correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = \frac{\text{"20"}}{\text{"20"}+5}\) or \(r = \frac{2\times\text{"20"}-24}{\text{"20"}}\) | M1 | Correct attempt at \(r\). Allow this to score for any of their \(k\) values. |
| \(r = \dfrac{4}{5}\) oe | A1 | Correct \(r\) from using \(k=20\). Allow this mark even if the 'other' value of \(r\) is also calculated. Allow unsimplified e.g. \(\dfrac{20}{20+5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = \text{"20"}+5 \Rightarrow S_{\infty} = \dfrac{\text{"25"}}{1-\text{"}\frac{4}{5}\text{"}}\) | M1 | Attempts to find \(a\) and \(S_{\infty}\) with \( |
| \(S_{\infty} = 125\) | A1 | Cao with no other values – if other values are found they must be clearly rejected and 125 "chosen". |
## Question 16:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2k-24}{k} = \frac{k}{k+5}$ or $\frac{k+5}{k} = \frac{k}{2k-24}$ or $(2k-24)(k+5) = k^2$ | M1 | Correct method. I.e. a method that uses the fact that the 3 terms are in geometric progression to establish an equation in $k$. |
| $(2k-24)(k+5) = 2k^2 - 14k - 120$ | dM1 | Expands $(2k-24)(k+5)$. Must be an attempt at the full expansion but allow the $k$ terms to be combined. **Dependent on the first M**. |
| $2k^2 - 14k - 120 = k^2 \Rightarrow k^2 - 14k - 120 = 0$* | A1* | Correct solution with no errors including bracketing errors e.g. $2k - 24(k+5) = \ldots$ |
**(3 marks)**
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(k+6)(k-20) = 0 \Rightarrow k = \ldots$ | M1 | Attempts to solve the given quadratic. See General Guidance. |
| $k = -6, \ 20$ | A1 | Correct values |
**(2 marks)**
**Part (c)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \frac{\text{"20"}}{\text{"20"}+5}$ or $r = \frac{2\times\text{"20"}-24}{\text{"20"}}$ | M1 | Correct attempt at $r$. Allow this to score for any of their $k$ values. |
| $r = \dfrac{4}{5}$ oe | A1 | Correct $r$ from using $k=20$. Allow this mark even if the 'other' value of $r$ is also calculated. Allow unsimplified e.g. $\dfrac{20}{20+5}$ |
**Part (c)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \text{"20"}+5 \Rightarrow S_{\infty} = \dfrac{\text{"25"}}{1-\text{"}\frac{4}{5}\text{"}}$ | M1 | Attempts to find $a$ and $S_{\infty}$ with $|r|<1$ |
| $S_{\infty} = 125$ | A1 | Cao with no other values – if other values are found they must be clearly rejected and 125 "chosen". |
**(4 marks)**
**Total: 9 marks**16. The first three terms of a geometric series are $( k + 5 ) , k$ and $( 2 k - 24 )$ respectively, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k ^ { 2 } - 14 k - 120 = 0$
\item Hence find the possible values of $k$.
\item Given that the series is convergent, find
\begin{enumerate}[label=(\roman*)]
\item the common ratio,
\item the sum to infinity.
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Leave blank & Q16 \\
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END & \\
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\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q16 [9]}}