| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Form and solve quadratic in parameter |
| Difficulty | Standard +0.3 This is a straightforward geometric sequence problem requiring students to use the constant ratio property to form a quadratic equation, solve it, then apply standard sum formulas. The algebraic manipulation is routine and the question clearly signposts each step, making it slightly easier than average for A-level. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If geometric, then \(r = \dfrac{3k-6}{5k-2}\) | M1 | Allow instead for \(r = \dfrac{k+2}{3k-6}\) or \(r^2 = \dfrac{k+2}{5k-2}\) in any form |
| Common ratio gives \(\dfrac{3k-6}{5k-2} = \dfrac{k+2}{3k-6}\) | M1 | Forms an equation in \(k\) which need not be simplified |
| So \(9k^2 - 36k + 36 = 5k^2 + 8k - 4\) | ||
| So \(k^2 - 11k + 10 = 0\) | A1 | AG — rearranges to correct three-term quadratic. At least one intermediate step must be shown. SC1 for showing \(k=1\) leads to \(3, -3, 3\) (\(r=-1\)) and \(k=10\) leads to \(48, 24, 12\) (\(r=\frac{1}{2}\)) and demonstrating both are geometric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| So \(k = 1, 10\) | M1 | Solves the quadratic to give at least one root |
| When \(k = 1\) the sum of 20 terms is \((3+(-3))+(3+(-3))+\cdots+(3+(-3)) = 0\) | M1 A1 | Evaluating the terms of the sequence when \(k=1\); cao |
| Alternative: \(S_{20} = \dfrac{3(1-(-1)^{20})}{1-(-1)} = 0\) | M1 A1 | Using the formula for sum of terms of a GP with \(r = -1\); cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(k = 10\) the sequence is \(48, 24, 12\ldots\) So \(a = 48,\ r = \dfrac{1}{2}\) | B1 | Identifies the first term and common ratio |
| \(S_\infty = \dfrac{48}{1 - \frac{1}{2}} = 96\) | B1 | cao |
## Question 11:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| If geometric, then $r = \dfrac{3k-6}{5k-2}$ | M1 | Allow instead for $r = \dfrac{k+2}{3k-6}$ or $r^2 = \dfrac{k+2}{5k-2}$ in any form |
| Common ratio gives $\dfrac{3k-6}{5k-2} = \dfrac{k+2}{3k-6}$ | M1 | Forms an equation in $k$ which need not be simplified |
| So $9k^2 - 36k + 36 = 5k^2 + 8k - 4$ | | |
| So $k^2 - 11k + 10 = 0$ | A1 | AG — rearranges to correct three-term quadratic. At least one intermediate step must be shown. **SC1** for showing $k=1$ leads to $3, -3, 3$ ($r=-1$) and $k=10$ leads to $48, 24, 12$ ($r=\frac{1}{2}$) and demonstrating both are geometric |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| So $k = 1, 10$ | M1 | Solves the quadratic to give at least one root |
| When $k = 1$ the sum of 20 terms is $(3+(-3))+(3+(-3))+\cdots+(3+(-3)) = 0$ | M1 A1 | Evaluating the terms of the sequence when $k=1$; cao |
| **Alternative:** $S_{20} = \dfrac{3(1-(-1)^{20})}{1-(-1)} = 0$ | M1 A1 | Using the formula for sum of terms of a GP with $r = -1$; cao |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $k = 10$ the sequence is $48, 24, 12\ldots$ So $a = 48,\ r = \dfrac{1}{2}$ | B1 | Identifies the first term and common ratio |
| $S_\infty = \dfrac{48}{1 - \frac{1}{2}} = 96$ | B1 | cao |11 The first three terms of a geometric sequence are $5 k - 2,3 k - 6 , k + 2$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k$ satisfies the equation $k ^ { 2 } - 11 k + 10 = 0$.
\item When $k$ takes the smaller of the two possible values, find the sum of the first 20 terms of the sequence.
\item When $k$ takes the larger of the two possible values, find the sum to infinity of the sequence.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2024 Q11 [8]}}