| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Prove term relationship |
| Difficulty | Standard +0.3 This is a structured multi-part question on geometric progressions with clear scaffolding. Part (i) requires translating a word statement into algebra (ar³ - a = 4(ar² - ar)), which simplifies straightforwardly. Parts (ii)-(iii) involve standard polynomial factorization techniques taught in C2. Part (iv) applies the sum to infinity formula. While it requires multiple steps, each part guides students through the process with no novel insights needed, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(ar^3 - a = 4(ar^2 - ar)\) | M1 | For using \(ar^{n-1}\) to form an equation |
| Hence \(r^3 - 4r^2 + 4r - 1 = 0\) | A1 | 2 For showing the given equation correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - 4 + 4 - 1 = 0\) | B1 | For correct substitution of \(r=1\), or state no remainder |
| Factors are \((r-1)(r^2-3r+1)\) | M1 | For attempted division, or equivalent |
| A1 | 3 For correct factor \(r^2 - 3r + 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \dfrac{3\pm\sqrt{5}}{2}\) | M1 | For solving the relevant quadratic equation |
| A1 | 2 For correct roots in exact form |
| Answer | Marks | Guidance |
|---|---|---|
| The relevant value of \(r\) is \(\dfrac{3-\sqrt{5}}{2}\) (or decimal equiv) | B1 | For selecting the appropriate value of \(r\) |
| Hence \(S_\infty = \dfrac{a}{1-\frac{1}{2}(3-\sqrt{5})}\) | M1 | For relevant use of \(\dfrac{a}{1-r}\) |
| \(= \dfrac{2a}{-1+\sqrt{5}} = \dfrac{2a(-1-\sqrt{5})}{(-1+\sqrt{5})(-1-\sqrt{5})}\) | M1 | For correct process for rationalising, using two term surd expression |
| \(= \frac{1}{2}a(1+\sqrt{5})\) | A1 | 4 For showing the given answer correctly |
# Question 9:
## Part (i):
| $ar^3 - a = 4(ar^2 - ar)$ | M1 | For using $ar^{n-1}$ to form an equation |
|---|---|---|
| Hence $r^3 - 4r^2 + 4r - 1 = 0$ | A1 | **2** For showing the given equation correctly |
## Part (ii):
| $1 - 4 + 4 - 1 = 0$ | B1 | For correct substitution of $r=1$, or state no remainder |
|---|---|---|
| Factors are $(r-1)(r^2-3r+1)$ | M1 | For attempted division, or equivalent |
| | A1 | **3** For correct factor $r^2 - 3r + 1$ |
## Part (iii):
| $r = \dfrac{3\pm\sqrt{5}}{2}$ | M1 | For solving the relevant quadratic equation |
|---|---|---|
| | A1 | **2** For correct roots in exact form |
## Part (iv):
| The relevant value of $r$ is $\dfrac{3-\sqrt{5}}{2}$ (or decimal equiv) | B1 | For selecting the appropriate value of $r$ |
|---|---|---|
| Hence $S_\infty = \dfrac{a}{1-\frac{1}{2}(3-\sqrt{5})}$ | M1 | For relevant use of $\dfrac{a}{1-r}$ |
| $= \dfrac{2a}{-1+\sqrt{5}} = \dfrac{2a(-1-\sqrt{5})}{(-1+\sqrt{5})(-1-\sqrt{5})}$ | M1 | For correct process for rationalising, using two term surd expression |
| $= \frac{1}{2}a(1+\sqrt{5})$ | A1 | **4** For showing the given answer correctly |9 A geometric progression has first term $a$, where $a \neq 0$, and common ratio $r$, where $r \neq 1$. The difference between the fourth term and the first term is equal to four times the difference between the third term and the second term.\\
(i) Show that $r ^ { 3 } - 4 r ^ { 2 } + 4 r - 1 = 0$.\\
(ii) Show that $r - 1$ is a factor of $r ^ { 3 } - 4 r ^ { 2 } + 4 r - 1$. Hence factorise $r ^ { 3 } - 4 r ^ { 2 } + 4 r - 1$.\\
(iii) Hence find the two possible values for the ratio of the geometric progression. Give your answers in an exact form.\\
(iv) For the value of $r$ for which the progression is convergent, prove that the sum to infinity is $\frac { 1 } { 2 } a ( 1 + \sqrt { } 5 )$.
\hfill \mbox{\textit{OCR C2 2005 Q9 [11]}}