| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Show quadratic equation in n |
| Difficulty | Moderate -0.8 This is a straightforward arithmetic sequence question with standard bookwork applications. Parts (a)-(c) involve direct formula substitution (finding terms and sums), part (d) is a 'show that' requiring simple algebraic manipulation of the sum formula, and part (e) solves a quadratic equation. All steps are routine C1 techniques with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(60 + 75 + 90 = 225^*\) or \(S_3 = \frac{3}{2}(120 + (3-1)(15)) = 225^*\) | B1* | Finds and adds the first 3 terms or uses sum of 3 terms of an AP and obtains the printed answer, with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(t_9 = 60 + (n-1)15 = (\pounds)180\) | M1 A1 | M1: Uses \(60 + (n-1)15\) with \(n = 8\) or \(9\). A1: \((\pounds)180\). Listing: M1: Uses \(a = 60\) and \(d = 15\) to select the \(8^{th}\) or \(9^{th}\) term (allow arithmetic slips). A1: \((\pounds)180\). (Special case \((\pounds)165\) only scores M1A0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_n = \frac{n}{2}(120 + (n-1)(15))\) or \(S_n = \frac{n}{2}(60 + 60 + (n-1)(15))\) | M1 | Uses correct formula for sum of \(n\) terms with \(a = 60\) and \(d = 15\) (must be a correct formula but ignore the value they use for \(n\) or could be in terms of \(n\)) |
| \(S_n = \frac{12}{2}(120 + (12-1)(15))\) | A1 | Correct numerical expression |
| \(= (\pounds)1710\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3375 = \frac{n}{2}(120 + (n-1)(15))\) | M1 | Uses correct formula for sum of \(n\) terms with \(a = 60\), \(d = 15\) and puts \(= 3375\) |
| \(6750 = 15n(8 + (n-1)) \Rightarrow 15n^2 + 105n = 6750\) | A1 | Correct three term quadratic. E.g. \(6750 = 105n + 15n^2\), \(3375 = \frac{15}{2}n^2 + \frac{105}{2}n\). This may be implied by equations such as \(6750 = 15n(n+7)\) or \(3375 = \frac{15}{2}(n^2 + 7n)\) |
| \(n^2 + 7n = 25 \times 18^*\) | A1* | Achieves the printed answer with no errors but must see the \(450\) or \(450\) in factorised form or e.g. \(6750\), \(3375\) in factorised form i.e. an intermediate step |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(n = 18 \Rightarrow\) Aged \(27\) | M1 A1 | M1: Attempts to solve the given quadratic or states \(n = 18\). A1: Age \(= 27\) or just \(27\). Age \(= 27\) only scores both marks (i.e. \(n = 18\) need not be seen) |
# Question 9:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $60 + 75 + 90 = 225^*$ or $S_3 = \frac{3}{2}(120 + (3-1)(15)) = 225^*$ | B1* | Finds and adds the first 3 terms or uses sum of 3 terms of an AP and obtains the printed answer, with no errors |
**Beware: The $12^{th}$ term of the sequence is 225 also so look out for $60 + (12-1) \times 15 = 225$. This is B0.**
**[1]**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t_9 = 60 + (n-1)15 = (\pounds)180$ | M1 A1 | M1: Uses $60 + (n-1)15$ with $n = 8$ or $9$. A1: $(\pounds)180$. Listing: M1: Uses $a = 60$ and $d = 15$ to select the $8^{th}$ or $9^{th}$ term (allow arithmetic slips). A1: $(\pounds)180$. (Special case $(\pounds)165$ **only** scores M1A0) |
**[2]**
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_n = \frac{n}{2}(120 + (n-1)(15))$ or $S_n = \frac{n}{2}(60 + 60 + (n-1)(15))$ | M1 | Uses correct formula for sum of $n$ terms with $a = 60$ and $d = 15$ (must be a correct formula but ignore the value they use for $n$ or could be in terms of $n$) |
| $S_n = \frac{12}{2}(120 + (12-1)(15))$ | A1 | Correct numerical expression |
| $= (\pounds)1710$ | A1 | cao |
**Listing: M1: Uses $a = 60$ and $d = 15$ and finds the sum of at least 12 terms (allow arithmetic slips). A2: $(\pounds)1710$**
**[3]**
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3375 = \frac{n}{2}(120 + (n-1)(15))$ | M1 | Uses correct formula for sum of $n$ terms with $a = 60$, $d = 15$ and puts $= 3375$ |
| $6750 = 15n(8 + (n-1)) \Rightarrow 15n^2 + 105n = 6750$ | A1 | **Correct three term** quadratic. E.g. $6750 = 105n + 15n^2$, $3375 = \frac{15}{2}n^2 + \frac{105}{2}n$. This may be implied by equations such as $6750 = 15n(n+7)$ or $3375 = \frac{15}{2}(n^2 + 7n)$ |
| $n^2 + 7n = 25 \times 18^*$ | A1* | Achieves the printed answer with no errors but must see the $450$ or $450$ in factorised form or e.g. $6750$, $3375$ in factorised form i.e. an intermediate step |
**[3]**
## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $n = 18 \Rightarrow$ Aged $27$ | M1 A1 | M1: Attempts to solve the given quadratic or states $n = 18$. A1: Age $= 27$ or just $27$. Age $= 27$ only scores both marks (i.e. $n = 18$ need not be seen) |
**Note that (e) is not hence so allow valid attempts to solve the given equation for M1**
**[2]**
**[11 marks total]**9. On John's 10th birthday he received the first of an annual birthday gift of money from his uncle. This first gift was $\pounds 60$ and on each subsequent birthday the gift was $\pounds 15$ more than the year before. The amounts of these gifts form an arithmetic sequence.
\begin{enumerate}[label=(\alph*)]
\item Show that, immediately after his 12th birthday, the total of these gifts was $\pounds 225$
\item Find the amount that John received from his uncle as a birthday gift on his 18th birthday.
\item Find the total of these birthday gifts that John had received from his uncle up to and including his 21st birthday.
When John had received $n$ of these birthday gifts, the total money that he had received from these gifts was $\pounds 3375$
\item Show that $n ^ { 2 } + 7 n = 25 \times 18$
\item Find the value of $n$, when he had received $\pounds 3375$ in total, and so determine John's age at this time.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2016 Q9 [11]}}