| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Largest or extreme value of sum |
| Difficulty | Moderate -0.3 This is a standard C1 arithmetic series question requiring systematic application of formulas (a+2d=5.5, S₄=22.75) to find a and d, then solving an inequality for when terms become negative. While it has multiple parts and requires careful algebraic manipulation with mixed numbers, it follows a predictable template with no novel problem-solving required—slightly easier than average due to its routine nature. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(a + 2d = 5\frac{1}{2}\) | (1) | B1 |
| \(\frac{3}{4}(2a + 3d) = 22\frac{1}{4}\) | (2) | M1 A1 |
| \((2) \Rightarrow 4a + 6d = 22\frac{3}{4}\) | ||
| \((1) \Rightarrow 3a + 6d = 16\frac{1}{2}\) | ||
| subtracting, \(a = 22\frac{3}{4} - 16\frac{1}{2} = 6\frac{1}{4}\) | M1 A1 | |
| \(d = \frac{1}{2}(5\frac{1}{2} - 6\frac{1}{4}) = -\frac{3}{8}\) | M1 A1 | |
| (b) \(6\frac{1}{4} - \frac{3}{8}(n - 1) > 0\) | M1 | |
| \(50 - 3(n - 1) > 0\) | ||
| \(n < 17\frac{2}{3} \therefore 17\) positive terms | M1 A1 | |
| (c) \(= S_{17} = \frac{17}{2}[12\frac{1}{4} + (16 \times -\frac{3}{8})]\) | M1 | |
| \(= \frac{17}{2}(12\frac{1}{2} - 6) = \frac{17}{2} \times \frac{13}{2} = \frac{221}{4} = 55\frac{1}{4}\) | A1 | (12) |
**(a)** $a + 2d = 5\frac{1}{2}$ | (1) | B1 |
$\frac{3}{4}(2a + 3d) = 22\frac{1}{4}$ | (2) | M1 A1 |
$(2) \Rightarrow 4a + 6d = 22\frac{3}{4}$ |
$(1) \Rightarrow 3a + 6d = 16\frac{1}{2}$ |
subtracting, $a = 22\frac{3}{4} - 16\frac{1}{2} = 6\frac{1}{4}$ | M1 A1 |
$d = \frac{1}{2}(5\frac{1}{2} - 6\frac{1}{4}) = -\frac{3}{8}$ | M1 A1 |
**(b)** $6\frac{1}{4} - \frac{3}{8}(n - 1) > 0$ | M1 |
$50 - 3(n - 1) > 0$ |
$n < 17\frac{2}{3} \therefore 17$ positive terms | M1 A1 |
**(c)** $= S_{17} = \frac{17}{2}[12\frac{1}{4} + (16 \times -\frac{3}{8})]$ | M1 |
$= \frac{17}{2}(12\frac{1}{2} - 6) = \frac{17}{2} \times \frac{13}{2} = \frac{221}{4} = 55\frac{1}{4}$ | A1 | (12)The third term of an arithmetic series is $5\frac{1}{2}$.
The sum of the first four terms of the series is $22\frac{3}{4}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the first term of the series is $6\frac{1}{4}$ and find the common difference. [7]
\item Find the number of positive terms in the series. [3]
\item Hence, find the greatest value of the sum of the first $n$ terms of the series. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q9 [12]}}