| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Two related arithmetic progressions |
| Difficulty | Standard +0.3 This is a straightforward multi-part arithmetic sequence question requiring standard formulas (nth term and sum). Part (a) is routine algebra with two equations, part (b) is direct substitution, and part (c) involves equating two sum formulas and solving a linear equation. All techniques are standard C1 content with no novel insight required, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(a + d = 26\) | M1 | |
| \(a + 4d = 41\) | A1 | |
| subtracting, \(3d = 15\) | M1 | |
| \(d = 5\) | A1 | |
| (b) \(a = 21\) | B1 | |
| \(u_{12} = 21 + (11 \times 5) = 76\) | M1 A1 | |
| (c) \(\frac{n}{2}[42 + 5(n-1)] = \frac{n}{2}[-24 + 7(n-1)]\) | M1 A1 | |
| \(n(5n + 37) = n(7n - 31)\) | ||
| \(2n(n - 34) = 0\) | M1 | |
| \(n > 0\) ⇒ \(n = 34\) | A1 | (11 marks) |
**(a)** $a + d = 26$ | M1 |
$a + 4d = 41$ | A1 |
subtracting, $3d = 15$ | M1 |
$d = 5$ | A1 |
**(b)** $a = 21$ | B1 |
$u_{12} = 21 + (11 \times 5) = 76$ | M1 A1 |
**(c)** $\frac{n}{2}[42 + 5(n-1)] = \frac{n}{2}[-24 + 7(n-1)]$ | M1 A1 |
$n(5n + 37) = n(7n - 31)$ | |
$2n(n - 34) = 0$ | M1 |
$n > 0$ ⇒ $n = 34$ | A1 | (11 marks)\begin{enumerate}
\item The second and fifth terms of an arithmetic series are 26 and 41 repectively.\\
(a) Show that the common difference of the series is 5 .\\
(b) Find the 12th term of the series.
\end{enumerate}
Another arithmetic series has first term -12 and common difference 7 .\\
Given that the sums of the first $n$ terms of these two series are equal,\\
(c) find the value of $n$.\\
\hfill \mbox{\textit{Edexcel C1 Q9 [11]}}