| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Sum of first n terms |
| Difficulty | Moderate -0.8 This is a straightforward application of standard geometric and arithmetic progression formulas with no problem-solving required. Part (a) uses direct substitution into GP formulas (nth term and sum), while part (b) requires solving a quadratic equation from the AP sum formula—all routine procedures below average difficulty for A-level. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(u_6 = 7 \times (-2)^5 = 1792\) | M1 | Attempt \(u_6\) using \(ar^{b}\). Allow for \(7 \times -2^5\). Using \(r = 2\) will be marked as a misread. |
| Obtain 1792 | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(S_{15} = \frac{7[1-(-2)^{15}]}{1-(-2)} = 76,461\) | M1 | Attempt sum of GP using correct formula. Must be using correct formula, so denominator of \(1-2\) is M0 unless \(1 - r\) clearly seen previously. If \(n = 14\) used, then only mark as misread if no contradictory evidence seen – starting with \(S_{15} = ...\) implies error in using in formula so M0. |
| Obtain 76,461 | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(\frac{N}{2}(2 \times 7 + (N - 1) \times (-2)) = -2900\), \(N(16 - 2N) = -5800\), \(N^2 - 8N - 2900 = 0\), \((N - 58)(N + 50) = 0\), \(N = 58\) | B1 | State correct unsimplified \(S_N\). If \((n-1)d\) is written as \((N-1) - 2\), then give benefit of doubt and allow B1, even if misused in subsequent work (eg becomes \(N - 3\)), unless there is clearly an error in the formula used. |
| Equate attempt at \(S_N\) to -2900 and rearrange to f(N) = 0 | M1 | Must be attempt at \(S_N\) for an AP, so using \(u_9\) or GP formulae will be M0. M0 if \((N-1) - 2\) becomes \(N - 3\). To give M1 at least one of the two terms in the bracket must have been multiplied by -2. Can still get M1 if incorrect formula as long as recognisable, and is quadratic in \(N\). Expand brackets and collect all terms on one side of equation. |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain \(N^2 - 8N - 2900 = 0\) | A1 | Any equivalent form as long as f(N) = 0 (but condone 0 not being explicit). |
| Attempt to solve 3 term quadratic | M1 | Any valid method – as long as it has come from equating an attempt at \(S_N\) of an AP to -2900.. |
| Obtain 58 only | A1 | 5 |
**(a)(i)** $u_6 = 7 \times (-2)^5 = 1792$ | M1 | Attempt $u_6$ using $ar^{b}$. Allow for $7 \times -2^5$. Using $r = 2$ will be marked as a misread.
**Obtain 1792** | A1 | 2 | Condone brackets not being shown explicitly in working.
SR B2 for listing terms, as long as signs change. Need to stop at $u_9$ or draw attention to it in a longer list.
**(ii)** $S_{15} = \frac{7[1-(-2)^{15}]}{1-(-2)} = 76,461$ | M1 | Attempt sum of GP using correct formula. Must be using correct formula, so denominator of $1-2$ is M0 unless $1 - r$ clearly seen previously. If $n = 14$ used, then only mark as misread if no contradictory evidence seen – starting with $S_{15} = ...$ implies error in using in formula so M0.
**Obtain 76,461** | A1 | 2 | Condone brackets not being shown explicitly in working.
SR B2 for listing terms and then manually adding them.
**(b)** $\frac{N}{2}(2 \times 7 + (N - 1) \times (-2)) = -2900$, $N(16 - 2N) = -5800$, $N^2 - 8N - 2900 = 0$, $(N - 58)(N + 50) = 0$, $N = 58$ | B1 | State correct unsimplified $S_N$. If $(n-1)d$ is written as $(N-1) - 2$, then give benefit of doubt and allow B1, even if misused in subsequent work (eg becomes $N - 3$), unless there is clearly an error in the formula used.
**Equate attempt at $S_N$ to -2900 and rearrange to f(N) = 0** | M1 | Must be attempt at $S_N$ for an AP, so using $u_9$ or GP formulae will be M0. M0 if $(N-1) - 2$ becomes $N - 3$. To give M1 at least one of the two terms in the bracket must have been multiplied by -2. Can still get M1 if incorrect formula as long as recognisable, and is quadratic in $N$. Expand brackets and collect all terms on one side of equation.
Allow slips eg not dividing all terms in bracket by 2.
**Obtain $N^2 - 8N - 2900 = 0$** | A1 | Any equivalent form as long as f(N) = 0 (but condone 0 not being explicit).
**Attempt to solve 3 term quadratic** | M1 | Any valid method – as long as it has come from equating an attempt at $S_N$ of an AP to -2900..
**Obtain 58 only** | A1 | 5 | 58 must clearly be intended as only final answer – could be through underlining, circling or deleting other value for $N$. No need to see other value for $N$ – if seen, allow slips as long as factorisation / substitution into formula is correct.
58 from answer only or trial and improvement can get 5/5.
58 and -50 with no working is 4/5.
---7
\begin{enumerate}[label=(\alph*)]
\item The first term of a geometric progression is 7 and the common ratio is - 2 .
\begin{enumerate}[label=(\roman*)]
\item Find the ninth term.
\item Find the sum of the first 15 terms.
\end{enumerate}\item The first term of an arithmetic progression is 7 and the common difference is - 2 . The sum of the first $N$ terms is - 2900 . Find the value of $N$.
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2011 Q7 [9]}}