| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find N for S_∞ - S_N condition |
| Difficulty | Standard +0.3 This is a straightforward multi-part geometric series question requiring standard formula application: S_∞ = a/(1-r) to find r, then using the nth term formula to find a, and finally computing S_∞ - S_30. All steps are routine bookwork with no problem-solving insight needed, 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 |
|---|---|---|
| \(\frac{a}{1-r}=6a\) | M1 | Either \(\frac{a}{1-r}=6a\) or \(\frac{6a}{1-r}=a\) or \(\frac{6}{1-r}=1\) |
| \(r=\frac{5}{6}\) | A1* | cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left\{T_4=ar^3=62.5\Rightarrow\right\}\ a\left(\frac{5}{6}\right)^3=62.5\) | M1 | Correct statement using the 4th term. Do not accept \(a\left(\frac{5}{6}\right)^4=62.5\) |
| \(a=108\) | A1 | 108 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_\infty=6(\text{their }a)\) or \(\frac{\text{their }a}{1-\frac{5}{6}}\{=648\}\) | M1 | Correct method to find \(S_\infty\) |
| \(\left\{S_{30}=\right\}\frac{108\left(1-\left(\frac{5}{6}\right)^{30}\right)}{1-\frac{5}{6}}\{=645.2701573\ldots\}\) | M1 A1ft | M1: \(S_{30}=\frac{(\text{their }a)\left(1-\left(\frac{5}{6}\right)^{30}\right)}{1-\left(\frac{5}{6}\right)}\). Condone invisible brackets around 5/6. A1ft: correct follow through expression (follow through their \(a\)). Do not condone invisible brackets around 5/6 unless their evaluation or final answer implies they were intended |
| \(\{S_\infty-S_{30}\}=2.72984\ldots\) | A1 | awrt 2.73 |
# Question 2(a):
$S_\infty = 6a$
| $\frac{a}{1-r}=6a$ | M1 | Either $\frac{a}{1-r}=6a$ or $\frac{6a}{1-r}=a$ or $\frac{6}{1-r}=1$ |
| $r=\frac{5}{6}$ | A1* | cso |
**[2]**
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# Question 2(b):
| $\left\{T_4=ar^3=62.5\Rightarrow\right\}\ a\left(\frac{5}{6}\right)^3=62.5$ | M1 | Correct statement using the 4th term. Do not accept $a\left(\frac{5}{6}\right)^4=62.5$ |
| $a=108$ | A1 | 108 |
**[2]**
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# Question 2(c):
| $S_\infty=6(\text{their }a)$ or $\frac{\text{their }a}{1-\frac{5}{6}}\{=648\}$ | M1 | Correct method to find $S_\infty$ |
| $\left\{S_{30}=\right\}\frac{108\left(1-\left(\frac{5}{6}\right)^{30}\right)}{1-\frac{5}{6}}\{=645.2701573\ldots\}$ | M1 A1ft | M1: $S_{30}=\frac{(\text{their }a)\left(1-\left(\frac{5}{6}\right)^{30}\right)}{1-\left(\frac{5}{6}\right)}$. Condone invisible brackets around 5/6. A1ft: correct follow through expression (follow through their $a$). **Do not condone invisible brackets around 5/6 unless their evaluation or final answer implies they were intended** |
| $\{S_\infty-S_{30}\}=2.72984\ldots$ | A1 | awrt 2.73 |
**[4] [Total 8]**
---2. A geometric series has first term $a$, where $a \neq 0$, and common ratio $r$. The sum to infinity of this series is 6 times the first term of the series.
\begin{enumerate}[label=(\alph*)]
\item Show that $r = \frac { 5 } { 6 }$
Given that the fourth term of this series is 62.5
\item find the value of $a$,
\item find the difference between the sum to infinity and the sum of the first 30 terms, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q2 [8]}}