| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | First-Order Linear Recurrence Relations |
| Difficulty | Moderate -0.3 This is a straightforward first-order linear recurrence question requiring substitution of given terms to find constants, then applying the formula and finding a limit. All steps are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = 0.5\); ordinates at \(x = 0, 0.5, 1, 1.5, 2\) | M1 | Correct strip width |
| \(y\) values: \(1,\ \sqrt{8(0.125)+1},\ 3,\ \sqrt{8(3.375)+1},\ \sqrt{33}\) | ||
| \(= 1,\ 1.7321,\ 3,\ 5.1926,\ 5.7446\) | A1 | |
| \(\approx \frac{0.5}{2}[1 + 5.7446 + 2(1.7321+3+5.1926)]\) | M1 | |
| \(= 0.25[6.7446 + 19.8494] = 6.65\) | A1 | 3 s.f. |
| Answer | Marks |
|---|---|
| Stretch in the \(x\)-direction, scale factor \(\frac{1}{2}\) | B1, B1 |
| Answer | Marks |
|---|---|
| Translation \(\begin{bmatrix}2\\-0.7\end{bmatrix}\) means \(g(x) = \sqrt{(x-2)^3+1} - 0.7\) | M1, A1 |
| \(g(4) = \sqrt{8+1} - 0.7 = 3 - 0.7 = 2.3\) | A1 |
# Question 5:
**(a)**
$h = 0.5$; ordinates at $x = 0, 0.5, 1, 1.5, 2$ | M1 | Correct strip width
$y$ values: $1,\ \sqrt{8(0.125)+1},\ 3,\ \sqrt{8(3.375)+1},\ \sqrt{33}$ | |
$= 1,\ 1.7321,\ 3,\ 5.1926,\ 5.7446$ | A1 |
$\approx \frac{0.5}{2}[1 + 5.7446 + 2(1.7321+3+5.1926)]$ | M1 |
$= 0.25[6.7446 + 19.8494] = 6.65$ | A1 | 3 s.f.
**(b)**
Stretch in the $x$-direction, scale factor $\frac{1}{2}$ | B1, B1 |
**(c)**
Translation $\begin{bmatrix}2\\-0.7\end{bmatrix}$ means $g(x) = \sqrt{(x-2)^3+1} - 0.7$ | M1, A1 |
$g(4) = \sqrt{8+1} - 0.7 = 3 - 0.7 = 2.3$ | A1 |
---5 The $n$th term of a sequence is $u _ { n }$.\\
The sequence is defined by
$$u _ { n + 1 } = p u _ { n } + q$$
where $p$ and $q$ are constants.
The first three terms of the sequence are given by
$$u _ { 1 } = 200 \quad u _ { 2 } = 150 \quad u _ { 3 } = 120$$
\begin{enumerate}[label=(\alph*)]
\item Show that $p = 0.6$ and find the value of $q$.
\item Find the value of $u _ { 4 }$.
\item The limit of $u _ { n }$ as $n$ tends to infinity is $L$. Write down an equation for $L$ and hence find the value of $L$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 Q5}}