| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compound growth applications |
| Difficulty | Standard +0.3 This is a straightforward geometric sequence application with common ratio 3. Parts (i)-(ii) are direct GP formula application, part (iii) is a standard GP sum formula (given as 'show that'), parts (iv)-(v) involve solving equations/inequalities with logarithms. All techniques are routine for C2 level with clear scaffolding throughout the multi-part structure. Slightly above average due to length and the logarithm work, but no novel problem-solving required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04k Modelling with sequences: compound interest, growth/decay1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks |
|---|---|
| \(81\) | 1 |
| Answer | Marks |
|---|---|
| \((1x)3^{n-1}\) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| (GP with) \(a = 1\) and \(r = 3\) | M1 | or M1 for \(= 1+3+9+\ldots+3^{n-1}\) |
| Clear correct use of GP sum formula | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (A) \(6\) www | 2 | |
| (B) | 1 | M1 for \(364 = (3^n - 1)/2\) |
| Answer | Marks | Guidance |
|---|---|---|
| their (ii) \(> 900\) | M1ft | \(-1\) once for \(=\) or \(<\) seen: condone wrong letter / missing brackets / no base |
| \((y-1)\log 3 > \log 900\) | M1ft | |
| \(y - 1 > \log 900 \div \log 3\) | M1 | |
| \(y = 8\) cao | B1 |
## Question 4:
**Part i:**
$81$ | 1 |
**Total: 1**
**Part ii:**
$(1x)3^{n-1}$ | 1 |
**Total: 1**
**Part iii:**
(GP with) $a = 1$ and $r = 3$ | M1 | or M1 for $= 1+3+9+\ldots+3^{n-1}$
Clear correct use of GP sum formula | M1 |
**Total: 2**
**Part iv:**
(A) $6$ www | 2 |
(B) | 1 | M1 for $364 = (3^n - 1)/2$
**Total: 3**
**Part v:**
their (ii) $> 900$ | M1ft | $-1$ once for $=$ or $<$ seen: condone wrong letter / missing brackets / no base
$(y-1)\log 3 > \log 900$ | M1ft |
$y - 1 > \log 900 \div \log 3$ | M1 |
$y = 8$ cao | B1 |
**Total: 4**4 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]{f291e6e3-975e-4d1e-aab6-67308f305da2-2_517_1116_356_455}
\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 Q4 [11]}}