| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Convergence conditions |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric sequence formulas with clear scaffolding. Part (a) is routine calculation, (b) tests basic understanding of convergence (|r|<1), (c) applies the sum to infinity formula directly, and (d) requires solving an equation but with standard algebraic manipulation. The context is accessible and all parts follow predictable patterns for C1/C2 level, 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_6 = 8000\times(0.85)^5 = 3549.6 \approx 3550\) | M1, A1 [2] | M1: attempts \(u_6=8000\times(r)^5\) with \(r=0.85\) or \(85\%\) or \(1-0.15\) or \(1-15\%\); A1: completes proof, states \(u_6=8000\times(0.85)^5\) and shows answer is awrt 3549.6 or 3550 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| States \( | r | <1\) or \(0.85<1\) and makes no reference to terms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_\infty = \frac{a}{1-r} = \frac{8000}{1-0.85}\) = awrt 53333 or 53334 or \(\frac{160000}{3}\) | M1A1 [2] | M1: attempts \(S_\infty=\frac{8000}{1-r}\) with \(r=0.85\) oe; A1: \(\frac{8000}{1-0.85}\) with answer awrt 53333 or 53334 or \(\frac{160000}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses \(S_N = \frac{8000(1-0.85^N)}{1-0.85}\) | M1 | |
| \(\frac{8000(1-0.85^N)}{1-0.85}=40000 \Rightarrow 0.85^N=0.25\) | dM1 A1 | |
| \(N=\frac{\log 0.25}{\log 0.85}(=8.53) \Rightarrow N=9\) | M1 A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(S_N = \frac{8000(1-r^N)}{1-r}\) with \(r = 0.85\) and \(S_N = 40000\) | M1 | Condone \(r = 0.15\) |
| Rearranges \(\frac{8000(1-r^N)}{1-r} = 40000\) to \(r^N = k\) with \(r = 0.85\) or \(0.15\) | dM1 | |
| \(0.85^N = 0.25\) | A1 | |
| Uses logs to solve \(a^N = b\), \((a,b > 0)\), reaching \(N = \ldots\) | M1 | Must be correct method; if just answer from \(a^N = b\), look for 1 dp accuracy; can start from \(40000 = 8000 \times (r')^{N-1}\) but must proceed to \(N=\) |
| \(N = 9\) | A1 | cso |
# Question 14:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_6 = 8000\times(0.85)^5 = 3549.6 \approx 3550$ | M1, A1 [2] | M1: attempts $u_6=8000\times(r)^5$ with $r=0.85$ or $85\%$ or $1-0.15$ or $1-15\%$; A1: completes proof, states $u_6=8000\times(0.85)^5$ and shows answer is awrt 3549.6 or 3550 |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| States $|r|<1$ or $0.85<1$ **and makes no reference to terms** | B1 [1] | Allow $r<1$ or $-1<r<1$ and makes no reference to terms; do not allow if they give $r=0.15$; do not allow explanation based around terms e.g. $8000\times0.85^{n-1}\to 0$ |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_\infty = \frac{a}{1-r} = \frac{8000}{1-0.85}$ = awrt 53333 or 53334 or $\frac{160000}{3}$ | M1A1 [2] | M1: attempts $S_\infty=\frac{8000}{1-r}$ with $r=0.85$ oe; A1: $\frac{8000}{1-0.85}$ with answer awrt 53333 or 53334 or $\frac{160000}{3}$ |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses $S_N = \frac{8000(1-0.85^N)}{1-0.85}$ | M1 | |
| $\frac{8000(1-0.85^N)}{1-0.85}=40000 \Rightarrow 0.85^N=0.25$ | dM1 A1 | |
| $N=\frac{\log 0.25}{\log 0.85}(=8.53) \Rightarrow N=9$ | M1 A1 [5] | |
## Question (d) [Geometric Series]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $S_N = \frac{8000(1-r^N)}{1-r}$ with $r = 0.85$ and $S_N = 40000$ | M1 | Condone $r = 0.15$ |
| Rearranges $\frac{8000(1-r^N)}{1-r} = 40000$ to $r^N = k$ with $r = 0.85$ or $0.15$ | dM1 | |
| $0.85^N = 0.25$ | A1 | |
| Uses logs to solve $a^N = b$, $(a,b > 0)$, reaching $N = \ldots$ | M1 | Must be correct method; if just answer from $a^N = b$, look for 1 dp accuracy; can start from $40000 = 8000 \times (r')^{N-1}$ but must proceed to $N=$ |
| $N = 9$ | A1 | cso |
---\begin{enumerate}
\item A new mineral has been discovered and is going to be mined over a number of years.
\end{enumerate}
A model predicts that the mass of the mineral mined each year will decrease by $15 \%$ per year, so that the mass of the mineral mined each year forms a geometric sequence.
Given that the mass of the mineral mined during year 1 is 8000 tonnes,\\
(a) show that, according to the model, the mass of the mineral mined during year 6 will be approximately 3550 tonnes.
According to the model, there is a limit to the total mass of the mineral that can be mined.\\
(b) With reference to the geometric series, state why this limit exists.\\
(c) Calculate the value of this limit.
It is decided that a total mass of 40000 tonnes of the mineral is required. This is going to be mined from year 1 to year $N$ inclusive.\\
(d) Assuming the model, find the value of $N$.\\
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\hfill \mbox{\textit{Edexcel C12 2017 Q14 [10]}}