OCR MEI C2 2012 June — Question 11 10 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2012
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeRelationship between two GPs
DifficultyStandard +0.3 This is a standard geometric progression problem requiring simultaneous equations from given conditions (second term and sum to infinity), solving a quadratic, then comparing nth terms. All techniques are routine for C2 level with clear scaffolding through parts (i) and (ii), making it slightly easier than average.
Spec1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1

11 A geometric progression has first term \(a\) and common ratio \(r\). The second term is 6 and the sum to infinity is 25 .
  1. Write down two equations in \(a\) and \(r\). Show that one possible value of \(a\) is 10 and find the other possible value of \(a\). Write down the corresponding values of \(r\).
  2. Show that the ratio of the \(n\)th terms of the two geometric progressions found in part (i) can be written as \(2 ^ { n - 2 } : 3 ^ { n - 2 }\).

Question 11(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(ar = 6\) oeB1 must be in \(a\) and \(r\)
\(\frac{a}{1-r} = 25\) oeB1 must be in \(a\) and \(r\)
\(25 = \frac{a}{1 - \frac{6}{a}}\)M1 or \(\frac{6}{r} = 25(1-r)\); NB assuming \(a = 10\) earns M0
\(a^2 - 25a + 150\ [= 0]\)A1 or \(25r^2 - 25r + 6\ [= 0]\); all signs may be reversed
\(a = 10\) obtained from formula, factorising, Factor theorem or completing the squareA1 if M0, B1 for \(r = 0.4\) and \(0.6\) and B1 for \(a = 15\) by trial and improvement; mark to benefit of candidate
\(a = 15\)A1 \(a = 15\)
\(r = 0.4\) and \(0.6\)A1 \(a = \frac{6}{0.6} = 10\) oe
Question 11(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(10 \times (3/5)^{n-1}\) and \(15 \times (2/5)^{n-1}\) seenM1
\(15 \times 2^{n-1} : 10 \times 3^{n-1}\) or \(3 \times \frac{2^{n-1}}{5^{n-1}} : 2 \times \frac{3^{n-1}}{5^{n-1}}\)M1 may be implied by \(3 \times 2^{n-1} : 2 \times 3^{n-1}\); condone ratio reversed
\(3 \times 2^{n-1} : 2 \times 3^{n-1}\)A1 and completion to given answer www; condone ratio reversed
Mark Scheme Examples (June 2012, Paper 4752)
Example 7:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Graph showing curve with points \((-1.3, 4.3)\) and \((1.3, -4.3)\) marked, crossing x-axis at \(-\sqrt{5}\) and \(\sqrt{5}\)B1 \(3^{\text{rd}}\) B1 condone extreme ends ruled
Example 8:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Graph crossing x-axis at \(-2\), \(\sqrt{c}\), and \(\sqrt{5}\) with correct cubic shapeB0 \(3^{\text{rd}}\) B0 — doesn't meet x-axis 3 times
Example 9:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Graph crossing x-axis at \(-\sqrt{5}\), \(0\), and \(\sqrt{5}\) with point \((0, 10)\) markedB1 \(4^{\text{th}}\) B1 is earned in spite of the curve not being a cubic
Example 10:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(0 = x(x^2 - 5)\)
\(x^2 = 5\)
\(x = \sqrt{5} = 2.23\) or \(-2.23\)
Points \((0, -2.23)\), \((0, 0)\), \((0, 2.23)\) marked on graphB1 \(4^{\text{th}}\) B1 — x-intercepts: co-ordinates reversed but condone this as candidates who write \(-2.23\), \(2.23\) only would not be penalised
Example 11:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Graph with \(x\)-intercepts at \(-\sqrt{5}\), \(0\), \(\sqrt{5}\) and local maximum \((1.3, 7.8)\) marked, value \(7.8\) on y-axisB1 \(3^{\text{rd}}\) B0: incorrect orientation — \(4^{\text{th}}\) B1 earned
Example 12:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x = 0,\ y = 0\)
\(x = 2.13,\ y = -4.3\)
\(x_2 = -1.3,\ y = 4.3\)
Points \((-\sqrt{5}, 0)\), \((0, 0)\), \((\sqrt{5}, 0)\) on x-axis; \((-1.3, 4.3)\) and \((1.3, -4.3)\) as turning pointsB1 \(4^{\text{th}}\) B1 earned
## Question 11(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $ar = 6$ oe | B1 | must be in $a$ and $r$ |
| $\frac{a}{1-r} = 25$ oe | B1 | must be in $a$ and $r$ |
| $25 = \frac{a}{1 - \frac{6}{a}}$ | M1 | or $\frac{6}{r} = 25(1-r)$; NB assuming $a = 10$ earns M0 |
| $a^2 - 25a + 150\ [= 0]$ | A1 | or $25r^2 - 25r + 6\ [= 0]$; all signs may be reversed |
| $a = 10$ obtained from formula, factorising, Factor theorem or completing the square | A1 | if M0, B1 for $r = 0.4$ **and** $0.6$ and B1 for $a = 15$ by trial and improvement; mark to benefit of candidate |
| $a = 15$ | A1 | $a = 15$ |
| $r = 0.4$ and $0.6$ | A1 | $a = \frac{6}{0.6} = 10$ oe |

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## Question 11(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $10 \times (3/5)^{n-1}$ and $15 \times (2/5)^{n-1}$ seen | M1 | |
| $15 \times 2^{n-1} : 10 \times 3^{n-1}$ or $3 \times \frac{2^{n-1}}{5^{n-1}} : 2 \times \frac{3^{n-1}}{5^{n-1}}$ | M1 | may be implied by $3 \times 2^{n-1} : 2 \times 3^{n-1}$; condone ratio reversed |
| $3 \times 2^{n-1} : 2 \times 3^{n-1}$ | A1 | and completion to given answer www; condone ratio reversed |

# Mark Scheme Examples (June 2012, Paper 4752)

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## Example 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph showing curve with points $(-1.3, 4.3)$ and $(1.3, -4.3)$ marked, crossing x-axis at $-\sqrt{5}$ and $\sqrt{5}$ | B1 | $3^{\text{rd}}$ B1 condone extreme ends ruled |

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## Example 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph crossing x-axis at $-2$, $\sqrt{c}$, and $\sqrt{5}$ with correct cubic shape | B0 | $3^{\text{rd}}$ B0 — doesn't meet x-axis 3 times |

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## Example 9:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph crossing x-axis at $-\sqrt{5}$, $0$, and $\sqrt{5}$ with point $(0, 10)$ marked | B1 | $4^{\text{th}}$ B1 is earned in spite of the curve not being a cubic |

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## Example 10:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = x(x^2 - 5)$ | | |
| $x^2 = 5$ | | |
| $x = \sqrt{5} = 2.23$ or $-2.23$ | | |
| Points $(0, -2.23)$, $(0, 0)$, $(0, 2.23)$ marked on graph | B1 | $4^{\text{th}}$ B1 — x-intercepts: co-ordinates reversed but condone this as candidates who write $-2.23$, $2.23$ only would not be penalised |

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## Example 11:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph with $x$-intercepts at $-\sqrt{5}$, $0$, $\sqrt{5}$ and local maximum $(1.3, 7.8)$ marked, value $7.8$ on y-axis | B1 | $3^{\text{rd}}$ B0: incorrect orientation — $4^{\text{th}}$ B1 earned |

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## Example 12:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 0,\ y = 0$ | | |
| $x = 2.13,\ y = -4.3$ | | |
| $x_2 = -1.3,\ y = 4.3$ | | |
| Points $(-\sqrt{5}, 0)$, $(0, 0)$, $(\sqrt{5}, 0)$ on x-axis; $(-1.3, 4.3)$ and $(1.3, -4.3)$ as turning points | B1 | $4^{\text{th}}$ B1 earned |
11 A geometric progression has first term $a$ and common ratio $r$. The second term is 6 and the sum to infinity is 25 .\\
(i) Write down two equations in $a$ and $r$. Show that one possible value of $a$ is 10 and find the other possible value of $a$. Write down the corresponding values of $r$.\\
(ii) Show that the ratio of the $n$th terms of the two geometric progressions found in part (i) can be written as $2 ^ { n - 2 } : 3 ^ { n - 2 }$.

\hfill \mbox{\textit{OCR MEI C2 2012 Q11 [10]}}