| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Total over time period |
| Difficulty | Moderate -0.8 This is a straightforward geometric progression question requiring only standard formula application: finding the common ratio (r=0.8), calculating the 20th term, sum of 20 terms, and sum to infinity. All formulas are given in the formula booklet and the question involves no problem-solving or conceptual challenges beyond direct substitution into memorized formulas. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{2400}{3000} = 0.8\) | B1 | For the correct value of \(r\) |
| M1 | For correct use of \(ar^{n-1}\) | |
| A1 | 3 | For correct (integer) answer |
| Forecast for week 20 is \(3000 \times 0.8^{19} = 43\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3000(1 - 0.8^{20})}{1 - 0.8} = 14827\) | M1 | For correct use of \(\frac{a(1-r^n)}{1-r}\) |
| A1 | 2 | For correct answer (3sf is acceptable) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3000}{1 - 0.8} = 15000\) | M1 | For correct use of \(\frac{a}{1-r}\) |
| A1 | 2 | For correct answer |
| Total: 7 |
## Part (i)
$r = \frac{2400}{3000} = 0.8$ | B1 | For the correct value of $r$
| M1 | For correct use of $ar^{n-1}$
| A1 | 3 | For correct (integer) answer
Forecast for week 20 is $3000 \times 0.8^{19} = 43$ | |
## Part (ii)
$\frac{3000(1 - 0.8^{20})}{1 - 0.8} = 14827$ | M1 | For correct use of $\frac{a(1-r^n)}{1-r}$
| A1 | 2 | For correct answer (3sf is acceptable)
## Part (iii)
$\frac{3000}{1 - 0.8} = 15000$ | M1 | For correct use of $\frac{a}{1-r}$
| A1 | 2 | For correct answer
| **Total: 7** | |Records are kept of the number of copies of a certain book that are sold each week. In the first week after publication 3000 copies were sold, and in the second week 2400 copies were sold. The publisher forecasts future sales by assuming that the number of copies sold each week will form a geometric progression with first two terms 3000 and 2400. Calculate the publisher's forecasts for
\begin{enumerate}[label=(\roman*)]
\item the number of copies that will be sold in the 20th week after publication, [3]
\item the total number of copies sold during the first 20 weeks after publication, [2]
\item the total number of copies that will ever be sold. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C2 Q4 [7]}}