| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sum of first n terms |
| Difficulty | Moderate -0.3 Part (a) involves straightforward arithmetic sequence calculations (finding 6th term and sum of 10 terms) using standard formulas. Part (b) requires identifying GP parameters and applying sum to infinity formula, plus a logarithm manipulation that's slightly more challenging but still routine for C2 level. All techniques are standard textbook exercises with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.06f Laws of logarithms: addition, subtraction, power rules1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks |
|---|---|
| (a)(i) \(3 + 5(2) = 13\) counters | B1 |
| (a)(ii) \(S_{10} = \frac{10}{2}(6 + 18) = 120\) | M1 A1 |
| (b)(i) \(P_4 = \frac{1}{6}\times\left(\frac{5}{6}\right)^3 = \frac{125}{1296}\) | M1 A1 |
| (b)(ii) First term \(= \frac{1}{6}\), common ratio \(= \frac{5}{6}\) | B1 |
| \(S_\infty = \frac{\frac{1}{6}}{1 - \frac{5}{6}} = \frac{\frac{1}{6}}{\frac{1}{6}} = 1\) | M1 A1 |
| (b)(iii) \(\frac{1}{6}\left(\frac{5}{6}\right)^{n-1} < 0.001\) | M1 |
| \((n-1)\log\frac{5}{6} < \log 0.006\) | M1 |
| \(n > \frac{\log 0.006}{\log(5/6)} + 1\) | A1 |
| \(n > \frac{-2.2218...}{-0.07918...} + 1 \approx 29.06\), so least \(n = 30\) | A1 |
## Question 11:
**(a)(i)** $3 + 5(2) = 13$ counters | B1 |
**(a)(ii)** $S_{10} = \frac{10}{2}(6 + 18) = 120$ | M1 A1 |
**(b)(i)** $P_4 = \frac{1}{6}\times\left(\frac{5}{6}\right)^3 = \frac{125}{1296}$ | M1 A1 |
**(b)(ii)** First term $= \frac{1}{6}$, common ratio $= \frac{5}{6}$ | B1 |
$S_\infty = \frac{\frac{1}{6}}{1 - \frac{5}{6}} = \frac{\frac{1}{6}}{\frac{1}{6}} = 1$ | M1 A1 |
**(b)(iii)** $\frac{1}{6}\left(\frac{5}{6}\right)^{n-1} < 0.001$ | M1 |
$(n-1)\log\frac{5}{6} < \log 0.006$ | M1 |
$n > \frac{\log 0.006}{\log(5/6)} + 1$ | A1 |
$n > \frac{-2.2218...}{-0.07918...} + 1 \approx 29.06$, so least $n = 30$ | A1 |11
\begin{enumerate}[label=(\alph*)]
\item André is playing a game where he makes piles of counters. He puts 3 counters in the first pile. Each successive pile he makes has 2 more counters in it than the previous one.
\begin{enumerate}[label=(\roman*)]
\item How many counters are there in his sixth pile?
\item André makes ten piles of counters. How many counters has he used altogether?
\end{enumerate}\item In another game, played with an ordinary fair die and counters, Betty needs to throw a six to start.
The probability $\mathrm { P } _ { n }$ of Betty starting on her $n$th throw is given by
$$P _ { n } = \frac { 1 } { 6 } \times \left( \frac { 5 } { 6 } \right) ^ { n - 1 }$$
\begin{enumerate}[label=(\roman*)]
\item Calculate $\mathrm { P } _ { 4 }$. Give your answer as a fraction.
\item The values $\mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } , \mathrm { P } _ { 3 } , \ldots$ form an infinite geometric progression. State the first term and the common ratio of this progression.
Hence show that $\mathrm { P } _ { 1 } + \mathrm { P } _ { 2 } + \mathrm { P } _ { 3 } + \ldots = 1$.
\item Given that $\mathrm { P } _ { n } < 0.001$, show that $n$ satisfies the inequality
$$n > \frac { \log _ { 10 } 0.006 } { \log _ { 10 } \left( \frac { 5 } { 6 } \right) } + 1$$
Hence find the least value of $n$ for which $\mathrm { P } _ { n } < 0.001$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2007 Q11 [12]}}