| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find term or common difference |
| Difficulty | Moderate -0.8 This is a straightforward application of standard arithmetic and geometric progression formulas. Part (a) requires substituting into S_n = n/2[2a + (n-1)d] and solving for d, while part (b) uses S_∞ = a/(1-r) with the given second term. Both are direct, single-step algebraic manipulations with no conceptual challenges—easier than average A-level questions. |
| 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|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(d = 5\) | M1, A1, M1, A1 4 | Attempt \(S_{70}\). Obtain correct unsimplified expression. Equate attempt at \(S_{70}\) to 12915, and attempt to find \(d\). Obtain \(d = 5\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(d = 5\) | M1, A1, M1, A1 4 | Attempt to find \(d\) by first equating \(\frac{l}{2}(a+l)\) to 12915. Obtain \(l = 357\). Equate \(u_{70}\) to \(l\). Obtain \(d = 5\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{a}{1-r} = 9\) | B1, B1 | Correct statement for second term. Correct statement for sum to infinity. |
| \(\frac{-4}{r} = 9 - 9r\) or \(a = 9 - (9 \times \frac{-4}{a})\) | M1 | Attempt to eliminate either \(a\) or \(r\). |
| \(9r^2 - 9r - 4 = 0\) or \(a^2 - 9a - 36 = 0\) | A1 | Obtain correct equation (no algebraic denominators/brackets). |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{4}{3}, r = -\frac{1}{3}\) or \(a = -3, a = 12\) | M1 | Attempt solution of three term quadratic equation. |
| Hence \(r = -\frac{1}{3}\) | A1, A1 7 | Obtain at least \(r = -\frac{1}{3}\) (from correct working only). Obtain \(r = -\frac{1}{3}\) only (from correct working only). SR: answer only / T&I is B2 only. |
(a) $S_{70} = \frac{70}{2}[2(12) + (70-1)d]$
$35(24 + 69d) = 12915$
$d = 5$ | M1, A1, M1, A1 4 | Attempt $S_{70}$. Obtain correct unsimplified expression. Equate attempt at $S_{70}$ to 12915, and attempt to find $d$. Obtain $d = 5$.
**OR**
$\frac{70}{2}[l + l] = 12915$
$l = 357$
$12 + 69d = 357$
$d = 5$ | M1, A1, M1, A1 4 | Attempt to find $d$ by first equating $\frac{l}{2}(a+l)$ to 12915. Obtain $l = 357$. Equate $u_{70}$ to $l$. Obtain $d = 5$.
(b) $ar = -4$
$\frac{a}{1-r} = 9$ | B1, B1 | Correct statement for second term. Correct statement for sum to infinity.
$\frac{-4}{r} = 9 - 9r$ or $a = 9 - (9 \times \frac{-4}{a})$ | M1 | Attempt to eliminate either $a$ or $r$.
$9r^2 - 9r - 4 = 0$ or $a^2 - 9a - 36 = 0$ | A1 | Obtain correct equation (no algebraic denominators/brackets).
$(3r-4)(3r+1) = 0$ or $(a+3)(a-12) = 0$
$r = \frac{4}{3}, r = -\frac{1}{3}$ or $a = -3, a = 12$ | M1 | Attempt solution of three term quadratic equation.
Hence $r = -\frac{1}{3}$ | A1, A1 7 | Obtain at least $r = -\frac{1}{3}$ (from correct working only). Obtain $r = -\frac{1}{3}$ only (from correct working only). SR: answer only / T&I is B2 only.7
\begin{enumerate}[label=(\alph*)]
\item In an arithmetic progression, the first term is 12 and the sum of the first 70 terms is 12915 . Find the common difference.
\item In a geometric progression, the second term is - 4 and the sum to infinity is 9 . Find the common ratio.
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2007 Q7 [11]}}