| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Prove sum formula |
| Difficulty | Moderate -0.5 Part (a) is a standard bookwork proof of the geometric series formula that appears in every C2 textbook and is frequently examined. Parts (b)-(d) involve routine algebraic manipulation to find r, a, and S_∞ using standard formulas. While multi-part, each step follows directly from applying memorized formulas with no novel problem-solving required. Slightly easier than average due to the straightforward nature of all parts. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((S_n =)\ a + ar + (ar^2) + \ldots + ar^{n-1}\) and \(rS_n = ar + ar^2 + (ar^3) + \ldots + ar^n\) | M1 | Lists both sums; \(S_n\) may be omitted, \(rS_n\) must be stated. Last term must be \(ar^{n-1}\) or \(ar^n\) in first series |
| \(S_n - rS_n = a - ar^n\) | M1 | Subtracts series for \(rS\) from \(S\) to give RHS \(= \pm(a - ar^n)\) |
| \(S_n(1-r) = a(1-r^n)\) | dM1 | Factorises both sides correctly; must follow from previous M1 |
| \(S_n = \frac{a(1-r^n)}{(1-r)}\) | A1 | Completes proof with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Divides one term by other (either way) to give \(r^2 = \ldots\) then square roots to give \(r =\) | M1 | Or (Method 2): finds geometric mean i.e. 3.24 and divides one term by 3.24 or 3.24 by one term |
| \(r^2 = \frac{1.944}{5.4},\quad r = 0.6\) (ignore \(-0.6\)) | A1 | Any correct equivalent for \(r\), e.g. 3/5. Answer only is 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(5.4 \div r^2\) or \(1.944 \div r^4\) to give \(a = 15\) | M1, A1ft | A1ft: follow through their value of \(r\). Just \(a = 15\) with no wrong working implies M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(S = \frac{15}{1-0.6}\) to obtain 37.5 | M1A1, A1 | M1: states sum to infinity formula with values of \(a\) and \(r\) found earlier, provided \( |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(S_n =)\ a + ar + (ar^2) + \ldots + ar^{n-1}$ **and** $rS_n = ar + ar^2 + (ar^3) + \ldots + ar^n$ | M1 | Lists both sums; $S_n$ may be omitted, $rS_n$ must be stated. Last term must be $ar^{n-1}$ or $ar^n$ in first series |
| $S_n - rS_n = a - ar^n$ | M1 | Subtracts series for $rS$ from $S$ to give RHS $= \pm(a - ar^n)$ |
| $S_n(1-r) = a(1-r^n)$ | dM1 | Factorises both sides correctly; must follow from previous M1 |
| $S_n = \frac{a(1-r^n)}{(1-r)}$ | A1 | Completes proof with **no errors** seen |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Divides one term by other (either way) to give $r^2 = \ldots$ then square roots to give $r =$ | M1 | Or (Method 2): finds geometric mean i.e. 3.24 and divides one term by 3.24 or 3.24 by one term |
| $r^2 = \frac{1.944}{5.4},\quad r = 0.6$ (ignore $-0.6$) | A1 | Any correct equivalent for $r$, e.g. 3/5. Answer only is 2/2 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $5.4 \div r^2$ or $1.944 \div r^4$ to give $a = 15$ | M1, A1ft | A1ft: follow through their value of $r$. Just $a = 15$ with no wrong working implies M1A1 |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $S = \frac{15}{1-0.6}$ to obtain 37.5 | M1A1, A1 | M1: states sum to infinity formula with values of $a$ and $r$ found earlier, provided $|r| < 1$. A1: uses 15 and 0.6 (or 3/5). A1: 37.5 or exact equivalent |\begin{enumerate}
\item A geometric series is $a + a r + a r ^ { 2 } + \ldots$\\
(a) Prove that the sum of the first $n$ terms of this series is given by
\end{enumerate}
$$S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$$
The third and fifth terms of a geometric series are 5.4 and 1.944 respectively and all the terms in the series are positive.
For this series find,\\
(b) the common ratio,\\
(c) the first term,\\
(d) the sum to infinity.\\
\hfill \mbox{\textit{Edexcel C2 2012 Q9 [11]}}