| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Prove sum formula |
| Difficulty | Moderate -0.3 Part (a) is a standard bookwork proof of the geometric series formula that appears in every textbook. Parts (b) and (c) are straightforward applications requiring recognition that r=0.93 and calculation of geometric terms/series. While multi-part with real-world context, this requires only routine application of formulas with no problem-solving insight, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S = a + ar + ar^2 + \ldots + ar^{n-1}\) and \(rS = ar + ar^2 + ar^3 + \ldots + ar^n\) | M1 | Minimum of 3 terms, must include first and \(n\)th term. Condone \(S = a + ar + ar^2 + \ldots + ar^n\) and \(rS = ar + ar^2 + \ldots + ar^{n+1}\) |
| \(S - rS = a - ar^n\) | M1 | Subtracts either way. Special case: allow \(S - rS = a + ar^n\). Their \(S\) and \(rS\) must be different |
| \(\Rightarrow S(1-r) = a(1-r^n) \Rightarrow S = \dfrac{a(1-r^n)}{(1-r)}\) | dM1A1* | dM1: dependent on both previous M's, taking out common factor of \(S\). A1*: fully correct proof, no errors or omissions. Commas instead of \(+\)'s is an error. \(S = \dfrac{a(r^n-1)}{(r-1)}\) without reaching printed answer is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S = \dfrac{(a + ar + ar^2 + \ldots + ar^{n-1})(1-r)}{1-r}\) | M1 | Minimum 3 terms, includes first and \(n\)th, multiplies top and bottom by \(1-r\) |
| \(S = \dfrac{a + ar + ar^2 + \ldots + ar^{n-1} - ar - ar^2 - \ldots - ar^n}{1-r}\) | M1 | Expands top with minimum 3 terms in each, must include first and \(n\)th term |
| \(S = \dfrac{a(1-r^n)}{(1-r)}\) | dM1A1 | dM1: dependent on both M's, taking out factor of \(a\) on top. A1: fully correct, no errors or omissions. Commas instead of \(+\)'s is an error. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(U = 180 \times 0.93^n\) with \(n=4\) or \(5\) | M1 | Attempts \(U = 180 \times 0.93^n\) with \(n=4\) or \(5\). Accept \(U = 167.4 \times 0.93^n\) with \(n=3\) or \(4\). Allow 93% for 0.93 |
| \(U_5 = 180 \times (0.93)^5 = 125.2\) (litres) | A1* | Cso. Awrt 125.2. Allow 93% or \(1-7\%\) for 0.93 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(S_n = \dfrac{a(1-r^n)}{(1-r)}\) with any combination of \(n=20/21\), \(a=180/167.4\), \(r=0.93\) | M1 | Allow 93% for 0.93 |
| \(S = \dfrac{167.4(1-0.93^{20})}{(1-0.93)}\) or \(S = 180 \times \dfrac{0.93(1-0.93^{20})}{(1-0.93)}\) or \(S = \dfrac{180(1-0.93^{21})}{(1-0.93)} - 180\) | A1 | Correct numerical expression for the sum (may be implied by awrt 1831). Allow 93% or \(1-7\%\) for 0.93 |
| 1831 (litres) | A1 | 1831 only (ignore units). Do not isw here, so \(1831 \times 20 = \ldots\) scores A0 |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S = a + ar + ar^2 + \ldots + ar^{n-1}$ **and** $rS = ar + ar^2 + ar^3 + \ldots + ar^n$ | M1 | Minimum of 3 terms, must include first and $n$th term. Condone $S = a + ar + ar^2 + \ldots + ar^n$ and $rS = ar + ar^2 + \ldots + ar^{n+1}$ |
| $S - rS = a - ar^n$ | M1 | Subtracts either way. Special case: allow $S - rS = a + ar^n$. Their $S$ and $rS$ must be different |
| $\Rightarrow S(1-r) = a(1-r^n) \Rightarrow S = \dfrac{a(1-r^n)}{(1-r)}$ | dM1A1* | dM1: dependent on both previous M's, taking out common factor of $S$. A1*: fully correct proof, **no errors or omissions**. Commas instead of $+$'s is an error. $S = \dfrac{a(r^n-1)}{(r-1)}$ without reaching printed answer is A0 |
### Part (a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S = \dfrac{(a + ar + ar^2 + \ldots + ar^{n-1})(1-r)}{1-r}$ | M1 | Minimum 3 terms, includes first and $n$th, multiplies top and bottom by $1-r$ |
| $S = \dfrac{a + ar + ar^2 + \ldots + ar^{n-1} - ar - ar^2 - \ldots - ar^n}{1-r}$ | M1 | Expands top with minimum 3 terms in each, must include first and $n$th term |
| $S = \dfrac{a(1-r^n)}{(1-r)}$ | dM1A1 | dM1: dependent on both M's, taking out factor of $a$ on top. A1: fully correct, **no errors or omissions**. Commas instead of $+$'s is an error. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $U = 180 \times 0.93^n$ with $n=4$ or $5$ | M1 | Attempts $U = 180 \times 0.93^n$ with $n=4$ or $5$. Accept $U = 167.4 \times 0.93^n$ with $n=3$ or $4$. Allow 93% for 0.93 |
| $U_5 = 180 \times (0.93)^5 = 125.2$ (litres) | A1* | Cso. Awrt 125.2. Allow 93% or $1-7\%$ for 0.93 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $S_n = \dfrac{a(1-r^n)}{(1-r)}$ with any combination of $n=20/21$, $a=180/167.4$, $r=0.93$ | M1 | Allow 93% for 0.93 |
| $S = \dfrac{167.4(1-0.93^{20})}{(1-0.93)}$ or $S = 180 \times \dfrac{0.93(1-0.93^{20})}{(1-0.93)}$ or $S = \dfrac{180(1-0.93^{21})}{(1-0.93)} - 180$ | A1 | Correct numerical expression for the sum (may be implied by awrt 1831). Allow 93% or $1-7\%$ for 0.93 |
| 1831 (litres) | A1 | 1831 **only** (ignore units). Do not isw here, so $1831 \times 20 = \ldots$ scores A0 |
---14. A geometric series has a first term $a$ and a common ratio $r$.
\begin{enumerate}[label=(\alph*)]
\item Prove that the sum of the first $n$ terms of this series is given by
$$S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$$
A liquid is to be stored in a barrel.
Due to evaporation, the volume of the liquid in a barrel at the end of a year is $7 \%$ less than the volume at the start of the year.
At the start of the first year, a barrel is filled with 180 litres of the liquid.
\item Show that the amount of the liquid in this barrel at the end of 5 years is approximately 125.2 litres.
At the start of each year a new identical barrel is filled with 180 litres of the liquid so that, at the end of 20 years, there are 20 barrels containing varying amounts of the liquid.
\item Calculate the total amount of the liquid, to the nearest litre, in the 20 barrels at the end of 20 years.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q14 [9]}}