| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| 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 progression application requiring use of the nth term formula (part a), sum formula with logarithms to solve for N (part b), and finding a specific term (part c). While it involves multiple steps and logarithms, the setup is clear, the techniques are standard Core 1/2 material, and the question guides students through the parts systematically, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Assumes GP and uses/states \(r = 1.06\) | B1 | Implied by any GP formula or term-by-term \(\times1.06\) |
| \(u_8 = ar^7 = 12\times(1.06)^7\) | M1 | Uses correct formula with correct \(a\), \(r\), \(n\) |
| \(= 12\times(1.06)^7 = 18.04\) so approximately 18 | A1* | Obtains awrt 18.0 or 18.04 before writing \(\approx 18\) km |
| Total | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{12(1.06^N-1)}{1.06-1}=1200 \Rightarrow r^N = k,\ k>0\) | M1 | Uses correct sum formula with their \(r\) and 1200; proceeds to \(r^N=k\) |
| \((1.06)^N = 7\) | A1 | |
| \(N = \frac{\log(7)}{\log 1.06} = 33.395 \Rightarrow N = 34\) | M1, A1 | M1: correct log method for power equation; A1: \(N=34\) cao; cannot be scored via incorrect inequality |
| Total | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance on day \(N\) is \(1200 - \frac{12((1.06)^{N-1}-1)}{1.06-1}\) | M1 | Correct expression, or \(1200-\text{"1168"}\) via trial and improvement |
| \(= \text{awrt } 32\text{ km}\) | A1 | 31.88 km or awrt 32 km (do not need to state km) |
| Total | [2] |
# Question 9:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assumes GP and uses/states $r = 1.06$ | B1 | Implied by any GP formula or term-by-term $\times1.06$ |
| $u_8 = ar^7 = 12\times(1.06)^7$ | M1 | Uses correct formula with correct $a$, $r$, $n$ |
| $= 12\times(1.06)^7 = 18.04$ so approximately 18 | A1* | Obtains awrt 18.0 or 18.04 before writing $\approx 18$ km |
| **Total** | **[3]** | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{12(1.06^N-1)}{1.06-1}=1200 \Rightarrow r^N = k,\ k>0$ | M1 | Uses correct sum formula with their $r$ and 1200; proceeds to $r^N=k$ |
| $(1.06)^N = 7$ | A1 | |
| $N = \frac{\log(7)}{\log 1.06} = 33.395 \Rightarrow N = 34$ | M1, A1 | M1: correct log method for power equation; A1: $N=34$ cao; cannot be scored via incorrect inequality |
| **Total** | **[4]** | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance on day $N$ is $1200 - \frac{12((1.06)^{N-1}-1)}{1.06-1}$ | M1 | Correct expression, or $1200-\text{"1168"}$ via trial and improvement |
| $= \text{awrt } 32\text{ km}$ | A1 | 31.88 km or awrt 32 km (do not need to state km) |
| **Total** | **[2]** | |
---9. A cyclist aims to travel a total of 1200 km over a number of days.
He cycles 12 km on day 1\\
He increases the distance that he cycles each day by $6 \%$ of the distance cycled on the previous day, until he reaches the total of 1200 km .
\begin{enumerate}[label=(\alph*)]
\item Show that on day 8 he cycles approximately 18 km .
He reaches his total of 1200 km on day $N$, where $N$ is a positive integer.
\item Find the value of $N$.
The cyclist stops when he reaches 1200 km .
\item Find the distance that he cycles on day $N$. Give your answer to the nearest km .
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q9 [9]}}