| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| 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 application of geometric sequences with clear scaffolding. Part (a) is a simple 'show that' calculation using the GP formula, part (b) requires solving a basic inequality with logarithms, and part (c) applies the sum formula for a GP. All techniques are standard C2 content with no problem-solving insight required, making it 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 |
| \(U_4 = 6000 \times (1.015)^3 = 6274\) (tonnes) | M1A1* | Use \(ar^3\) with \(a=6000\), \(r=1.015\). Condone \(r=1.15\) or \(1.0015\). Accept a list of 4 terms with same conditions. cso \(6000\times(1.015)^3 = 6274\) tonnes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(U_N = 6000 \times 1.015^{N-1} > 8000\) | M1 | Attempts \(ar^{n-1}\ldots 8000\) or \(ar^n\ldots 8000\) with \(a=6000\), \(r=1.015\), condoning \(r\) being 1.15 or 1.0015 |
| \(1.015^{N-1} > \frac{8000}{6000}\) | A1 | For reaching \(1.015^{N-1}\ldots\frac{4}{3}\) or \(1.015^n\ldots\frac{4}{3}\). Allow \(\frac{4}{3}\) rounded or truncated to 1.33 (2dp or better) |
| \(\log(1.015^{N-1}) > \log\left(\frac{4}{3}\right) \Rightarrow N > \frac{\log\left(\frac{4}{3}\right)}{\log(1.015)}+1\ (=20.3)\) | M1A1 | Uses logs correctly to get \(n\) or \(n-1\). Correct (unrounded) answer, may be left in log form. If \(n\) used instead of \(n-1\), must subsequently reach \(N=21\) |
| \(N = 21\) | A1 | Do not accept \(N>21\) etc. Two final A marks may be implied by finding '\(n\)' and adding 1 to reach 21 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_n = \frac{a(1-r^n)}{1-r}\) with \(n=10\), \(a=6000/30000\), \(r=1.015\) | M1 | Accept \(a=6000/30000\), \(r=1.015/1+1.5\%/1.15/1.0015\). Alternatively accept list of ten terms starting £30000, £30450, £30906.75,... added or £6000, £6090, £6181.35,... added |
| \(S = 5\times\frac{6000(1.015^{10}-1)}{(1.015-1)}\) OR \(S = \frac{30000(1.015^{10}-1)}{(1.015-1)}\) | A1 | Correct unsimplified answer to the cost for 10 years |
| Awrt £321 000 | A1 | cso awrt £321 000 |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $U_4 = 6000 \times (1.015)^3 = 6274$ (tonnes) | M1A1* | Use $ar^3$ with $a=6000$, $r=1.015$. Condone $r=1.15$ or $1.0015$. Accept a list of 4 terms with same conditions. cso $6000\times(1.015)^3 = 6274$ tonnes |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $U_N = 6000 \times 1.015^{N-1} > 8000$ | M1 | Attempts $ar^{n-1}\ldots 8000$ or $ar^n\ldots 8000$ with $a=6000$, $r=1.015$, condoning $r$ being 1.15 or 1.0015 |
| $1.015^{N-1} > \frac{8000}{6000}$ | A1 | For reaching $1.015^{N-1}\ldots\frac{4}{3}$ or $1.015^n\ldots\frac{4}{3}$. Allow $\frac{4}{3}$ rounded or truncated to 1.33 (2dp or better) |
| $\log(1.015^{N-1}) > \log\left(\frac{4}{3}\right) \Rightarrow N > \frac{\log\left(\frac{4}{3}\right)}{\log(1.015)}+1\ (=20.3)$ | M1A1 | Uses logs correctly to get $n$ or $n-1$. Correct (unrounded) answer, may be left in log form. If $n$ used instead of $n-1$, must subsequently reach $N=21$ |
| $N = 21$ | A1 | Do not accept $N>21$ etc. Two final A marks may be implied by finding '$n$' and adding 1 to reach 21 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_n = \frac{a(1-r^n)}{1-r}$ with $n=10$, $a=6000/30000$, $r=1.015$ | M1 | Accept $a=6000/30000$, $r=1.015/1+1.5\%/1.15/1.0015$. Alternatively accept list of ten terms starting £30000, £30450, £30906.75,... added or £6000, £6090, £6181.35,... added |
| $S = 5\times\frac{6000(1.015^{10}-1)}{(1.015-1)}$ OR $S = \frac{30000(1.015^{10}-1)}{(1.015-1)}$ | A1 | Correct unsimplified answer to the cost for 10 years |
| Awrt £321 000 | A1 | cso awrt £321 000 |
---11. Wheat is to be grown on a farm.
A model predicts that the mass of wheat harvested on the farm will increase by $1.5 \%$ per year, so that the mass of wheat harvested each year forms a geometric sequence.
Given that the mass of wheat harvested during year one is 6000 tonnes,
\begin{enumerate}[label=(\alph*)]
\item show that, according to the model, the mass of wheat harvested on the farm during year 4 will be approximately 6274 tonnes.
During year $N$, according to the model, there is predicted to be more than 8000 tonnes of wheat harvested on the farm.
\item Find the smallest possible value of $N$.
It costs $\pounds 5$ per tonne to harvest the wheat.
\item Assuming the model, find the total amount that it would cost to harvest the wheat from year one to year 10 inclusive. Give your answer to the nearest $\pounds 1000$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q11 [10]}}