| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sum of multiples or integers |
| Difficulty | Moderate -0.8 This is a straightforward C1 arithmetic sequences question requiring only standard formula application: part (a) is direct sum formula with n=50, a=2, l=100; part (b) uses the same formula with algebraic manipulation; part (c) applies the nth term formula. All parts are routine textbook exercises with no problem-solving insight required, making it 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) Series has 50 terms | B1 | |
| \(S = \frac{1}{2}(50)(2 + 100) = 2550\) or \(S = \frac{1}{2}(50)(4 + 49 \times 2) = 2550\) | M1 A1 | |
| Notes | (a) B for seeing attempt to use \(n = 50\) or \(n = 50\) stated. M for attempt to use \(\frac{1}{2}n(a+l)\) or \(\frac{1}{2}n(2a + (n-1)d)\) with \(a = 2\) and values for other variables (Using \(n = 100\) may earn B0 M1A0). |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{100}{k}\) | B1 | |
| (ii) Sum: \(\frac{1}{2}\left(\frac{100}{k}\right)(k + 100)\) or \(\frac{1}{2}\left(\frac{100}{k}\right)\left(2k + \left(\frac{100}{k} - 1\right)k\right)\) | M1 A1 | |
| \(= 50 + \frac{5000}{k}\) | A1 cso | (*) |
| Notes | (b) M for use of \(a = k\) and \(d = k\) or \(l = 100\) with their value for n, could be numerical or even letter n in correct formula for sum. A1: Correct formula with \(n = 100/k\). A1: NB Answer is printed – so no slips should have appeared in working. | |
| (c) 50th term = \(a + (n - 1)d = (2k + 1) + 49"(2k + 3)"\) or \(= 100k + 148\) | M1 | |
| Or \(2k + 49(2k + 3) = 100k + 148\) | A1 | |
| Notes | (c) M for use of formula \(a + 49d\) with \(a = 2k + 1\) and with d obtained from difference of terms. A1: Requires this simplified answer. |
(a) Series has 50 terms | B1 |
$S = \frac{1}{2}(50)(2 + 100) = 2550$ or $S = \frac{1}{2}(50)(4 + 49 \times 2) = 2550$ | M1 A1 |
Notes | | (a) B for seeing attempt to use $n = 50$ or $n = 50$ stated. M for attempt to use $\frac{1}{2}n(a+l)$ or $\frac{1}{2}n(2a + (n-1)d)$ with $a = 2$ and values for other variables (Using $n = 100$ may earn B0 M1A0).
(b)
(i) $\frac{100}{k}$ | B1 |
(ii) Sum: $\frac{1}{2}\left(\frac{100}{k}\right)(k + 100)$ or $\frac{1}{2}\left(\frac{100}{k}\right)\left(2k + \left(\frac{100}{k} - 1\right)k\right)$ | M1 A1 |
$= 50 + \frac{5000}{k}$ | A1 cso | (*)
Notes | | (b) M for use of $a = k$ and $d = k$ or $l = 100$ with their value for n, could be numerical or even letter n in correct formula for sum. A1: Correct formula with $n = 100/k$. A1: NB Answer is printed – so no slips should have appeared in working.
(c) 50th term = $a + (n - 1)d = (2k + 1) + 49"(2k + 3)"$ or $= 100k + 148$ | M1 |
| Or $2k + 49(2k + 3) = 100k + 148$ | A1 |
Notes | | (c) M for use of formula $a + 49d$ with $a = 2k + 1$ and with d obtained from difference of terms. A1: Requires this simplified answer.
---\begin{enumerate}
\item (a) Calculate the sum of all the even numbers from 2 to 100 inclusive,
\end{enumerate}
$$2 + 4 + 6 + \ldots \ldots + 100$$
(b) In the arithmetic series
$$k + 2 k + 3 k + \ldots \ldots + 100$$
$k$ is a positive integer and $k$ is a factor of 100 .\\
(i) Find, in terms of $k$, an expression for the number of terms in this series.\\
(ii) Show that the sum of this series is
$$50 + \frac { 5000 } { k }$$
(c) Find, in terms of $k$, the 50th term of the arithmetic sequence
$$( 2 k + 1 ) , ( 4 k + 4 ) , ( 6 k + 7 ) , \ldots \ldots ,$$
giving your answer in its simplest form.\\
\hfill \mbox{\textit{Edexcel C1 2011 Q9 [9]}}