| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 8 |
| Topic | Geometric Sequences and Series |
| Type | Find N for S_∞ - S_N condition |
| Difficulty | Moderate -0.3 This is a straightforward geometric progression question requiring standard formulas and techniques. Part (i) is routine calculation of GP terms, part (ii) involves algebraic manipulation of the sum formula (but the inequality is given to show, not derive), and part (iii) is a standard logarithm application. While it requires multiple steps across three parts, each component uses well-practiced A-level techniques without requiring problem-solving insight or novel approaches. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{3}{4}\) | B1 | |
| \(u_5 = 12r^4\) | M1 | Applying their \(r\) in the correct formula for \(u_5\) with \(a = 12\); Or repeated use of their \(r\) |
| \(u_5 = \frac{243}{64}\) | A1 | 3.796 875 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_\infty = \frac{12}{1-\frac{3}{4}}\) or \(S_N = \frac{12(1-(\frac{3}{4})^N)}{1-\frac{3}{4}}\) | B1ft | Correctly applying formula for \(S_\infty\) or \(S_N\) with their value of \(r\) |
| \(\frac{12}{1-\frac{3}{4}} - \frac{12(1-(\frac{3}{4})^N)}{1-\frac{3}{4}} \leq 0.0096\) | M1 | Attempt at \(S_\infty - S_N\) compared with 0.0096 (dependent on previous B1); Accept any inequality or equals for this mark |
| \(48 - 48(1-(\frac{3}{4})^N) \leq 0.0096 \Rightarrow (\frac{3}{4})^N \leq 0.0002\) | A1 | AG — completely correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(N \log(\frac{3}{4}) \leq \log(0.0002) \Rightarrow N \geq \ldots\) | M1 | Take logs and attempt to make \(N\) the subject (accept any inequality or equals for this mark); \(N = \log_4(0.0002)\) |
| \(N \geq 29.6\ldots \Rightarrow N = 30\) | A1 |
## Part (i):
$r = \frac{3}{4}$ | B1 |
$u_5 = 12r^4$ | M1 | Applying their $r$ in the correct formula for $u_5$ with $a = 12$; Or repeated use of their $r$
$u_5 = \frac{243}{64}$ | A1 | 3.796 875
## Part (ii):
$S_\infty = \frac{12}{1-\frac{3}{4}}$ or $S_N = \frac{12(1-(\frac{3}{4})^N)}{1-\frac{3}{4}}$ | B1ft | Correctly applying formula for $S_\infty$ or $S_N$ with their value of $r$
$\frac{12}{1-\frac{3}{4}} - \frac{12(1-(\frac{3}{4})^N)}{1-\frac{3}{4}} \leq 0.0096$ | M1 | Attempt at $S_\infty - S_N$ compared with 0.0096 (dependent on previous B1); Accept any inequality or equals for this mark
$48 - 48(1-(\frac{3}{4})^N) \leq 0.0096 \Rightarrow (\frac{3}{4})^N \leq 0.0002$ | A1 | AG — completely correct working
## Part (iii):
$N \log(\frac{3}{4}) \leq \log(0.0002) \Rightarrow N \geq \ldots$ | M1 | Take logs and attempt to make $N$ the subject (accept any inequality or equals for this mark); $N = \log_4(0.0002)$
$N \geq 29.6\ldots \Rightarrow N = 30$ | A1 |The first term of a geometric progression is 12 and the second term is 9.
\begin{enumerate}[label=(\roman*)]
\item Find the fifth term. [3]
\end{enumerate}
The sum of the first $N$ terms is denoted by $S_N$ and the sum to infinity is denoted by $S_\infty$. It is given that the difference between $S_\infty$ and $S_N$ is at most 0.0096.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that $\left(\frac{3}{4}\right)^N \leqslant 0.0002$. [3]
\item Use logarithms to find the smallest possible value of $N$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q2 [8]}}