| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find first term from conditions |
| Difficulty | Moderate -0.3 This is a straightforward geometric progression question requiring standard formula application. Part (i) is a simple algebraic manipulation of S∞ = a/(1-r), part (ii) uses ar² = 9 with known r, and part (iii) applies the sum formula. All steps are routine with no problem-solving insight needed, 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/Working | Mark | Guidance |
| \(4a = \frac{a}{1-r}\) | M1 | Equate \(\frac{a}{1-r}\) to \(4a\), or substitute \(r = \frac{3}{4}\) into \(S_\infty\). \(S_\infty\) must be quoted correctly. Allow \(4ar^0\) for \(4a\). Initially using a numerical value for \(a\) is M0. |
| \(1 - r = \frac{1}{4}\) | M1 | Attempt to find value for \(r\) or evaluate \(S_\infty\). Need at least one extra line of working between initial statement and given answer. |
| \(r = \frac{3}{4}\) | A1 | 3 marks Obtain \(r = \frac{3}{4}\) (or show \(S_\infty = 4a\)). Allow \(r = 0.75\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{3}{4}\right)^2 a = 9\) | M1* | Attempt use of \(ar^2\). Must use \(r = \frac{3}{4}\). Must be clearly intended as \(ar^2\). |
| \(a = 16\) | M1d* | Equate to 9 and attempt to find \(a\). Must get as far as attempting value for \(a\). |
| \(a = 16\) | A1 | 3 marks Obtain \(a = 16\). Answer only gets full credit. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{20} = \frac{16\left(1 - \frac{3}{4}^{20}\right)}{1 - \frac{3}{4}}\) | M1 | Attempt use of correct sum formula for a GP. Must be correct formula, with \(a\) = their (ii), \(r = \frac{3}{4}\) and \(n = 20\). |
| \(= 63.8\) | A1 | 2 marks Obtain 63.8 or better. More accurate answer is 63.79704... NB using \(n-1\) rather than \(n\) gives 63.729 (M0), using \(n+1\) gives 63.848 (M0). Must be decimal rather than exact answer with power of \(\frac{3}{4}\). |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4a = \frac{a}{1-r}$ | M1 | Equate $\frac{a}{1-r}$ to $4a$, or substitute $r = \frac{3}{4}$ into $S_\infty$. $S_\infty$ must be quoted correctly. Allow $4ar^0$ for $4a$. Initially using a numerical value for $a$ is M0. |
| $1 - r = \frac{1}{4}$ | M1 | Attempt to find value for $r$ or evaluate $S_\infty$. Need at least one extra line of working between initial statement and given answer. |
| $r = \frac{3}{4}$ | A1 | **3 marks** Obtain $r = \frac{3}{4}$ (or show $S_\infty = 4a$). Allow $r = 0.75$. |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{3}{4}\right)^2 a = 9$ | M1* | Attempt use of $ar^2$. Must use $r = \frac{3}{4}$. Must be clearly intended as $ar^2$. |
| $a = 16$ | M1d* | Equate to 9 and attempt to find $a$. Must get as far as attempting value for $a$. |
| $a = 16$ | A1 | **3 marks** Obtain $a = 16$. Answer only gets full credit. |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{20} = \frac{16\left(1 - \frac{3}{4}^{20}\right)}{1 - \frac{3}{4}}$ | M1 | Attempt use of correct sum formula for a GP. Must be correct formula, with $a$ = their (ii), $r = \frac{3}{4}$ and $n = 20$. |
| $= 63.8$ | A1 | **2 marks** Obtain 63.8 or better. More accurate answer is 63.79704... NB using $n-1$ rather than $n$ gives 63.729 (M0), using $n+1$ gives 63.848 (M0). Must be decimal rather than exact answer with power of $\frac{3}{4}$. |
---5 In a geometric progression, the sum to infinity is four times the first term.\\
(i) Show that the common ratio is $\frac { 3 } { 4 }$.\\
(ii) Given that the third term is 9 , find the first term.\\
(iii) Find the sum of the first twenty terms.
\hfill \mbox{\textit{OCR C2 2011 Q5 [8]}}