| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find year when threshold exceeded |
| Difficulty | Standard +0.3 This is a straightforward geometric sequences question with common ratio 3. Parts (i)-(iv) involve direct application of GP formulas and solving 3^n = 729. Part (v) requires logarithms but follows a standard template for 'find the year when threshold exceeded' problems. All steps are routine for C2 level with no novel insights required. |
| 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 rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (i) 81 | 1 | |
| (ii) \((1x)3^{n-1}\) | 1 | |
| (iii) (GP with) \(a = 1\) and \(r = 3\) clear correct use GP sum formula | M1 M1 | or M1 for \(= 1+3+9+\ldots+3^{n-1}\) |
| (iv) (A) 6 www (B) 243 | M1 for \(364 = (3^n - 1)/2\) | 3 |
| (v) their (ii) \(> 900\) \((y - 1)\log 3 > \log 900\) \(y - 1 > \log 900 \div \log 3\) \(y = 8\) cao | M1ft M1ft M1 B1 | \(-1\) once for \(=\) or \(<\) seen: condone wrong letter / missing brackets / no base |
(i) 81 | 1
(ii) $(1x)3^{n-1}$ | 1
(iii) (GP with) $a = 1$ and $r = 3$ clear correct use GP sum formula | M1 M1 | or M1 for $= 1+3+9+\ldots+3^{n-1}$ | 2
(iv) (A) 6 www (B) 243 | M1 for $364 = (3^n - 1)/2$ | 3
(v) their (ii) $> 900$ $(y - 1)\log 3 > \log 900$ $y - 1 > \log 900 \div \log 3$ $y = 8$ cao | M1ft M1ft M1 B1 | $-1$ once for $=$ or $<$ seen: condone wrong letter / missing brackets / no base | 411 There is a flowerhead at the end of each stem of an oleander plant. The next year, each flowerhead is replaced by three stems and flowerheads, as shown in Fig. 11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{faeaf2aa-ed4e-4926-b402-40c4c9aad479-5_501_1102_431_504}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item How many flowerheads are there in year 5?
\item How many flowerheads are there in year $n$ ?
\item As shown in Fig. 11, the total number of stems in year 2 is 4, (that is, 1 old one and 3 new ones). Similarly, the total number of stems in year 3 is 13 , (that is, $1 + 3 + 9$ ).
Show that the total number of stems in year $n$ is given by $\frac { 3 ^ { n } - 1 } { 2 }$.
\item Kitty's oleander has a total of 364 stems. Find\\
(A) its age,\\
(B) how many flowerheads it has.
\item Abdul's oleander has over 900 flowerheads.
Show that its age, $y$ years, satisfies the inequality $y > \frac { \log _ { 10 } 900 } { \log _ { 10 } 3 } + 1$.\\
Find the smallest integer value of $y$ for which this is true.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2005 Q11 [10]}}