| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find year when threshold exceeded |
| Difficulty | Moderate -0.3 This is a straightforward geometric sequence/series question requiring standard formula application: (a) uses nth term formula, (b) uses sum formula, and (c) requires solving 300(1.05)^(N-1) = 3000 using logarithms. All techniques are routine for C2 level with no novel problem-solving required, 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 |
|---|---|---|
| Uses \(300 \times (1.05)^{23}\); obtains 921 or 922 or 920 | M1, A1 | cao; this answer implies the M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \(S = \frac{300(1.05^{24}-1)}{1.05-1}\); obtains 13351 (accept awrt 13400) | M1, A1 | Must have correct \(r\) and \(n\) but can use their \(a\); answers which round to 13400 acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \(300(1.05)^{n-1} > 3000\) or \(300(1.05)^{n-1} = 3000\) | M1 | Correct inequality or equality, interpreted correctly later |
| \((n-1)\log 1.05 > \log 10\) or \((n-1)\log 1.05 = \log 10\) or \((n-1) = \log_{1.05} 10\) | M1 | Correct algebra then correct use of logs |
| \(n > 48.19\), \(N = 49\) | A1 | Need to see 49 or 49th month; trial and error: 49 is M1M1A1, 48 scores M1M0A0 |
## Question 9:
### Part (a):
| Uses $300 \times (1.05)^{23}$; obtains 921 or 922 or 920 | M1, A1 | cao; this answer implies the M1 |
### Part (b):
| Uses $S = \frac{300(1.05^{24}-1)}{1.05-1}$; obtains 13351 (accept awrt 13400) | M1, A1 | Must have correct $r$ and $n$ but can use their $a$; answers which round to 13400 acceptable |
### Part (c):
| Uses $300(1.05)^{n-1} > 3000$ or $300(1.05)^{n-1} = 3000$ | M1 | Correct inequality or equality, interpreted correctly later |
| $(n-1)\log 1.05 > \log 10$ or $(n-1)\log 1.05 = \log 10$ or $(n-1) = \log_{1.05} 10$ | M1 | Correct algebra then correct use of logs |
| $n > 48.19$, $N = 49$ | A1 | Need to see 49 or 49th month; trial and error: 49 is M1M1A1, 48 scores M1M0A0 |
---9. In the first month after opening, a mobile phone shop sold 300 phones. A model for future sales assumes that the number of phones sold will increase by $5 \%$ per month, so that $300 \times 1.05$ will be sold in the second month, $300 \times 1.05 ^ { 2 }$ in the third month, and so on.
Using this model, calculate
\begin{enumerate}[label=(\alph*)]
\item the number of phones sold in the 24th month,
\item the total number of phones sold over the whole 24 months.
This model predicts that, in the $N$ th month, the number of phones sold in that month exceeds 3000 for the first time.
\item Find the value of $N$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q9 [7]}}