| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Relationship between two GPs |
| Difficulty | Standard +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. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
## 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 |
---
## 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 |
---
## 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 |
---
## 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 |
---
## 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 |
---
## 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 |
---
## 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]}}