| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find year when threshold exceeded |
| Difficulty | Standard +0.3 This is a straightforward geometric sequence application with standard steps: verifying a calculation (part a), identifying the common ratio (part b), manipulating a logarithmic inequality (part c), and using a calculator to find N (part d). All techniques are routine for C1/C2 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(130000 \times (1.02) = 132600\) or \(2\% = 2600\) and \(130000 + 2600 = 132600\) | B1 | A reason must be provided; allow \(130000\times(1+2\%)\) and \(130000\times(102\%)\); but not \(130000\times1+2\%\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = 1.02\) | B1 | For 1.02; allow e.g. \(\frac{51}{50}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(130000\times(1.02)^{N-1} > 260000\) or \(= 260000\) | M1 | Correct inequality or equality; may use \(r\) or 1.02; may use \(N\) or \(n\) |
| \((1.02)^{N-1} > 2\) | A1 | cao; allow \((1.02)^{N-1} > 2\) |
| \((N-1)\log_{10}(1.02) > \log_{10}2\) or \((N-1)\log_{10}(1.02) = \log_{10}2\) or \((N-1) > \log_{1.02}2\) | M1 | Correct use of logs power rule on previous line from \(n\)th term of GP; condone missing brackets; allow base absent or 'ln' |
| \(N > \frac{\log_{10}2}{\log_{10}(1.02)} + 1\) | A1cso | Answer exactly as printed including bases; all inequality work correct; no missing brackets earlier |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N = 37\) | B1 | Only need \(N=37\); may follow trial and error or logs to different base; do not allow \(N\geq37\) or \(N>37\) or \(N=37.0\) |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $130000 \times (1.02) = 132600$ or $2\% = 2600$ and $130000 + 2600 = 132600$ | B1 | A reason must be provided; allow $130000\times(1+2\%)$ and $130000\times(102\%)$; but not $130000\times1+2\%$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = 1.02$ | B1 | For 1.02; allow e.g. $\frac{51}{50}$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $130000\times(1.02)^{N-1} > 260000$ or $= 260000$ | M1 | Correct inequality or equality; may use $r$ or 1.02; may use $N$ or $n$ |
| $(1.02)^{N-1} > 2$ | A1 | cao; allow $(1.02)^{N-1} > 2$ |
| $(N-1)\log_{10}(1.02) > \log_{10}2$ or $(N-1)\log_{10}(1.02) = \log_{10}2$ or $(N-1) > \log_{1.02}2$ | M1 | Correct use of logs power rule on previous line from $n$th term of GP; condone missing brackets; allow base absent or 'ln' |
| $N > \frac{\log_{10}2}{\log_{10}(1.02)} + 1$ | A1cso | Answer exactly as printed including bases; all inequality work correct; no missing brackets earlier |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N = 37$ | B1 | Only need $N=37$; may follow trial and error or logs to different base; do not allow $N\geq37$ or $N>37$ or $N=37.0$ |
---9. The resident population of a city is 130000 at the end of Year 1
A model predicts that the resident population of the city will increase by $2 \%$ each year, with the populations at the end of each year forming a geometric sequence.
\begin{enumerate}[label=(\alph*)]
\item Show that the predicted resident population at the end of Year 2 is 132600
\item Write down the value of the common ratio of the geometric sequence.
The model predicts that Year $N$ will be the first year which will end with the resident population of the city exceeding 260000
\item Show that
$$N > \frac { \log _ { 10 } 2 } { \log _ { 10 } 1.02 } + 1$$
\item Find the value of $N$.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q9 [7]}}