| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Prove sum formula |
| Difficulty | Moderate -0.8 Part (a) is a standard bookwork proof of the geometric series formula that appears in every A-level textbook, requiring only algebraic manipulation of Sn - rSn. Part (b) involves substituting the formula and solving a simple equation with r^5, which is routine application. This is easier than average as it's mostly recall and standard technique with no novel problem-solving required. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_n = a + ar + ar^2 + \ldots + ar^{n-1}\) | B1 | States correct series |
| \(rS_n = ar + ar^2 + ar^3 + \ldots + ar^n \Rightarrow S_n - rS_n = \ldots\) | M1 | Subtracts to obtain \(S_n - rS_n\) |
| \(S_n - rS_n = a - ar^n\) | A1 | Correct result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_n(1-r) = a(1-r^n) \Rightarrow S_n = \dfrac{a(1-r^n)}{(1-r)}\) | A1* | Depends on all previous marks; must show all steps with both sides unfactorised |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{a(1-r^{10})}{1-r} = 4 \times \dfrac{a(1-r^5)}{1-r}\) or \(4 \times \dfrac{a(1-r^{10})}{1-r} = \dfrac{a(1-r^5)}{1-r}\) | M1 | Equation in \(r^{10}\) and \(r^5\) (and possibly \(1-r\)); must cancel through by \(a\) |
| \(1 - r^{10} = 4(1-r^5)\) | A1 | Correct equation with \(a\) and \(1-r\) cancelled |
| \(r^{10} - 4r^5 + 3 = 0 \Rightarrow (r^5-1)(r^5-3)=0 \Rightarrow r^5 = \ldots\) or \(1-r^{10}=4(1-r^5)\Rightarrow(1-r^5)(1+r^5)=4(1-r^5)\Rightarrow r^5=\ldots\) | dM1 | Depends on first M; solve 3-term quadratic in \(r^5\) by valid method |
| \(r = \sqrt[5]{3}\) oe only | A1 | \(r=1\) if found must be rejected |
| Total: (4) |
## Question 15(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_n = a + ar + ar^2 + \ldots + ar^{n-1}$ | B1 | States correct series |
| $rS_n = ar + ar^2 + ar^3 + \ldots + ar^n \Rightarrow S_n - rS_n = \ldots$ | M1 | Subtracts to obtain $S_n - rS_n$ |
| $S_n - rS_n = a - ar^n$ | A1 | Correct result |
## Question 15 (Geometric Series):
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_n(1-r) = a(1-r^n) \Rightarrow S_n = \dfrac{a(1-r^n)}{(1-r)}$ | A1* | Depends on all previous marks; must show all steps with both sides unfactorised |
| **Total: (4)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{a(1-r^{10})}{1-r} = 4 \times \dfrac{a(1-r^5)}{1-r}$ or $4 \times \dfrac{a(1-r^{10})}{1-r} = \dfrac{a(1-r^5)}{1-r}$ | M1 | Equation in $r^{10}$ and $r^5$ (and possibly $1-r$); must cancel through by $a$ |
| $1 - r^{10} = 4(1-r^5)$ | A1 | Correct equation with $a$ and $1-r$ cancelled |
| $r^{10} - 4r^5 + 3 = 0 \Rightarrow (r^5-1)(r^5-3)=0 \Rightarrow r^5 = \ldots$ or $1-r^{10}=4(1-r^5)\Rightarrow(1-r^5)(1+r^5)=4(1-r^5)\Rightarrow r^5=\ldots$ | dM1 | Depends on first M; solve 3-term quadratic in $r^5$ by valid method |
| $r = \sqrt[5]{3}$ oe only | A1 | $r=1$ if found must be rejected |
| **Total: (4)** | | |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
A geometric series has common ratio $r$ and first term $a$.\\
Given $r \neq 1$ and $a \neq 0$\\
(a) prove that
$$S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$$
Given also that $S _ { 10 }$ is four times $S _ { 5 }$\\
(b) find the exact value of $r$.
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q15 [8]}}