| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: evaluate sum |
| Difficulty | Moderate -0.8 This is a straightforward C1 question requiring basic substitution into a recurrence relation to work backwards (finding a₁ from a₂), then forward to generate terms, followed by simple addition. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial since it requires careful arithmetic across multiple steps. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks |
|---|---|
| (a) \(7 = 5a_1 - 3 \Rightarrow a_1 = \ldots\) | M1 |
| \(a_1 = 2\) | A1 |
| (2 marks) | |
| (b) \(a_3 = \text{"32"}\) and \(a_4 = \text{"157"}\) | M1 |
| \(\sum_{r=1}^{4} a_r = a_1 + a_2 + a_3 + a_4\) | dM1 |
| \(= \text{"2"}+ \text{"7"}+ \text{"32"}+ \text{"157"}\) | |
| \(= 198\) | A1 |
| (3 marks) | |
| Notes: | |
| (a) | M1: Writes \(7 = 5a_1 - 3\) and attempts to solve leading to an answer for \(a_1\). If they rearrange wrongly before any substitution this is M0. A1: Cao \(a_1 = 2\). Special case: Substitutes \(n = 1\) into \(5n - 3\) and obtains answer \(2\). This is fortuitous and gets M0A0 but full marks are available on (b). |
| (b) | M1: Attempts to find either their \(a_3\) or their \(a_4\) using \(a_{n+1} = 5a_n - 3\), \(a_2 = 7\). Needs clear attempt to use formula or is implied by correct answers or correct follow through of their \(a_3\). dM1: (Depends on previous M mark) Sum of their four adjacent terms from the given sequence. n.b May be given for \(9 + a_3 + a_4\) as they may add \(2 + 7\) to give \(9\) (dM0 for sum of an Arithmetic series). A1: cao \(198\). |
| Special case: | \((a) a_1 = 32\) is M0 A0. (b) Adds for example \(7+32+157 + 782 = \quad\) or \(32+157 + 782 + 3907\) is M1 M1 A0. Total mark possible is \(2 / 5\) (This is not treated as a misread – as it changes the question). |
**(a)** $7 = 5a_1 - 3 \Rightarrow a_1 = \ldots$ | M1 |
$a_1 = 2$ | A1 |
| | **(2 marks)** |
**(b)** $a_3 = \text{"32"}$ and $a_4 = \text{"157"}$ | M1 |
$\sum_{r=1}^{4} a_r = a_1 + a_2 + a_3 + a_4$ | dM1 |
$= \text{"2"}+ \text{"7"}+ \text{"32"}+ \text{"157"}$ | |
$= 198$ | A1 |
| | **(3 marks)** |
| **Notes:** | |
|---|---|
| **(a)** | M1: Writes $7 = 5a_1 - 3$ and attempts to solve leading to an answer for $a_1$. If they rearrange wrongly before any substitution this is M0. A1: Cao $a_1 = 2$. Special case: Substitutes $n = 1$ into $5n - 3$ and obtains answer $2$. This is fortuitous and gets M0A0 but full marks are available on (b). |
| **(b)** | M1: Attempts to find either their $a_3$ or their $a_4$ using $a_{n+1} = 5a_n - 3$, $a_2 = 7$. Needs clear attempt to use formula or is implied by correct answers or correct follow through of their $a_3$. dM1: (Depends on previous M mark) Sum of their four adjacent terms from the given sequence. n.b May be given for $9 + a_3 + a_4$ as they may add $2 + 7$ to give $9$ (dM0 for sum of an Arithmetic series). A1: cao $198$. |
| **Special case:** | $(a) a_1 = 32$ is M0 A0. (b) Adds for example $7+32+157 + 782 = \quad$ or $32+157 + 782 + 3907$ is M1 M1 A0. Total mark possible is $2 / 5$ (This is not treated as a misread – as it changes the question). |
---5. A sequence of numbers $a _ { 1 } , a _ { 2 } , a _ { 3 } \ldots$ is defined by
$$a _ { n + 1 } = 5 a _ { n } - 3 , \quad n \geqslant 1$$
Given that $a _ { 2 } = 7$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $a _ { 1 }$
\item Find the value of $\sum _ { r = 1 } ^ { 4 } a _ { r }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q5 [5]}}