| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Arithmetic progression with parameters |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing basic definitions of arithmetic and geometric progressions. Part (i) requires setting up one equation using the constant difference property (routine algebra). Part (ii) involves verifying a GP condition with a given value and finding common ratio, then solving a quadratic equation. All techniques are standard C2 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x+4) - 2x = (2x-7) - (x+4)\) OR \(2x + d = x + 4\) and \(2x + 2d = 2x - 7\) | M1 | Attempt to eliminate \(d\) to obtain equation in \(x\) only. Equate two expressions for \(d\), both in terms of \(x\) |
| Correct equation in just \(x\) | A1 | Allow unsimplified equation. A0 if brackets missing unless implied by subsequent working |
| \(x = 7.5\) | A1 | Any equivalent form. Allow from no working or T&I. Alt method: B1 state \(2x + 2d = 2x - 7\) to obtain \(d = -3.5\); M1 attempt to find \(x\); A1 obtain \(x = 7.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Terms are \(16, 12, 9\) | B1 | Ignore any additional terms given |
| \(\frac{12}{16} = 0.75\), \(\frac{9}{12} = 0.75\), common ratio of \(0.75\) so GP | B1 | Must show two values of \(0.75\), or unsimplified fractions. Must state ratio has been checked twice using both pairs of consecutive terms. SR B2 if \(16, 12, 9\) never stated explicitly but used convincingly with \(r = 0.75\) twice |
| \(S_\infty = \frac{16}{1-0.75} = 64\) | M1 | Attempt use of \(\frac{a}{1-r}\). Must be correct formula. Allow with incorrect \(a\) and/or \(r\) |
| \(= 64\) | A1 | A0 if given as 'approximately 64' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{(2x-7)}{(x+4)} = \frac{(x+4)}{2x}\), giving \(4x^2 - 14x = x^2 + 8x + 16\) | M1* | Attempt to eliminate \(r\) to obtain equation in \(x\) only. Equate two expressions for \(r\), both in terms of \(x\) |
| \(3x^2 - 22x - 16 = 0\) | A1 | Allow \(6x^2 - 44x - 32 = 0\). Allow any equivalent form as long as like terms combined |
| \((3x+2)(x-8) = 0\) | M1d* | Attempt to solve quadratic. Dependent on first M1 |
| \(x = -\frac{2}{3}\) | A1 | Allow recurring decimal, but not rounded or truncated. Condone \(x = 8\) also given |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x+4) - 2x = (2x-7) - (x+4)$ **OR** $2x + d = x + 4$ and $2x + 2d = 2x - 7$ | M1 | Attempt to eliminate $d$ to obtain equation in $x$ only. Equate two expressions for $d$, both in terms of $x$ |
| Correct equation in just $x$ | A1 | Allow unsimplified equation. A0 if brackets missing unless implied by subsequent working |
| $x = 7.5$ | A1 | Any equivalent form. Allow from no working or T&I. **Alt method:** B1 state $2x + 2d = 2x - 7$ to obtain $d = -3.5$; M1 attempt to find $x$; A1 obtain $x = 7.5$ |
### Part (ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Terms are $16, 12, 9$ | B1 | Ignore any additional terms given |
| $\frac{12}{16} = 0.75$, $\frac{9}{12} = 0.75$, common ratio of $0.75$ so GP | B1 | Must show two values of $0.75$, or unsimplified fractions. Must state ratio has been checked twice using both pairs of consecutive terms. **SR B2** if $16, 12, 9$ never stated explicitly but used convincingly with $r = 0.75$ twice |
| $S_\infty = \frac{16}{1-0.75} = 64$ | M1 | Attempt use of $\frac{a}{1-r}$. Must be correct formula. Allow with incorrect $a$ and/or $r$ |
| $= 64$ | A1 | A0 if given as 'approximately 64' |
### Part (ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(2x-7)}{(x+4)} = \frac{(x+4)}{2x}$, giving $4x^2 - 14x = x^2 + 8x + 16$ | M1* | Attempt to eliminate $r$ to obtain equation in $x$ only. Equate two expressions for $r$, both in terms of $x$ |
| $3x^2 - 22x - 16 = 0$ | A1 | Allow $6x^2 - 44x - 32 = 0$. Allow any equivalent form as long as like terms combined |
| $(3x+2)(x-8) = 0$ | M1d* | Attempt to solve quadratic. Dependent on first M1 |
| $x = -\frac{2}{3}$ | A1 | Allow recurring decimal, but not rounded or truncated. Condone $x = 8$ also given |
---6 (i) The first three terms of an arithmetic progression are $2 x , x + 4$ and $2 x - 7$ respectively. Find the value of $x$.\\
(ii) The first three terms of another sequence are also $2 x , x + 4$ and $2 x - 7$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Verify that when $x = 8$ the terms form a geometric progression and find the sum to infinity in this case.
\item Find the other possible value of $x$ that also gives a geometric progression.
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2013 Q6 [11]}}