Question 9 - A-Level Maths

Edexcel C1 2005 June — Question 9 13 marks

Exam Board	Edexcel
Module	C1 (Core Mathematics 1)
Year	2005
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Arithmetic Sequences and Series
Type	Prove sum formula
Difficulty	Moderate -0.8 This is a straightforward C1 question testing standard arithmetic series knowledge. Part (a) is a bookwork proof that should be memorized. Parts (b)-(e) involve routine substitution into formulas with simple arithmetic and solving a quadratic equation. The context is accessible and all steps are procedural with no problem-solving insight required.
Spec	1.01a Proof: structure of mathematical proof and logical steps 1.02f Solve quadratic equations: including in a function of unknown 1.04h Arithmetic sequences: nth term and sum formulae 1.04i Geometric sequences: nth term and finite series sum

9. An arithmetic series has first term $a$ and common difference $d$.

Prove that the sum of the first $n$ terms of the series is $$\frac { 1 } { 2 } n [ 2 a + ( n - 1 ) d ] .$$ Sean repays a loan over a period of $n$ months. His monthly repayments form an arithmetic sequence. He repays $\pounds 149$ in the first month, $\pounds 147$ in the second month, $\pounds 145$ in the third month, and so on. He makes his final repayment in the $n$th month, where $n > 21$.
Find the amount Sean repays in the 21st month. Over the $n$ months, he repays a total of $\pounds 5000$.
Form an equation in $n$, and show that your equation may be written as $$n ^ { 2 } - 150 n + 5000 = 0 .$$
Solve the equation in part (c).
State, with a reason, which of the solutions to the equation in part (c) is not a sensible solution to the repayment problem.

Show mark scheme Show mark scheme source

Question 9:

Part (a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$S=a+(a+d)+\ldots+[a+(n-1)d]$	B1	Requires at least 3 terms, must include first and last terms, an adjacent term, dots and $+$ signs
$S=[a+(n-1)d]+\ldots+a$	M1	For reversing series; must be arithmetic with $a$, $d$ (or $a$, $l$) and $n$
$2S=[2a+(n-1)d]+\ldots+[2a+(n-1)d]$	dM1	For adding; must have $2S$ and be genuine attempt; dependent on 1st M1
$S=\frac{n}{2}[2a+(n-1)d]$	A1 c.s.o.	Allow first 3 marks for use of $l$ for last term but as given for final mark

Part (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$a=149$, $d=-2$; $u_{21}=149+20(-2)=\pounds109$	M1 A1	For using $a=149$ and $d=\pm 2$ in $a+(n-1)d$ formula

Part (c):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$S_n=\frac{n}{2}[2\times149+(n-1)(-2)]$ $(=n(150-n))$	M1 A1	For using their $a$, $d$ in $S_n$; A1 any correct expression
$S_n=5000 \Rightarrow n^2-150n+5000=0$	A1 c.s.o.	For putting $S_n=5000$ and simplifying to given expression; no wrong work

Part (d):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$(n-100)(n-50)=0$	M1	Attempt to solve leading to $n=\ldots$
$n=50$ or $100$	A2/1/0	Give A1A0 for 1 correct value; A1A1 for both correct

Part (e):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$u_{100}<0$ $\therefore n=100$ not sensible	B1 f.t.	Must mention 100 and state $u_{100}<0$ (or loan paid or equivalent); if giving f.t. then must have $n\geq 76$

## Question 9:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S=a+(a+d)+\ldots+[a+(n-1)d]$ | B1 | Requires at least 3 terms, must include first and last terms, an adjacent term, dots and $+$ signs |
| $S=[a+(n-1)d]+\ldots+a$ | M1 | For reversing series; must be arithmetic with $a$, $d$ (or $a$, $l$) and $n$ |
| $2S=[2a+(n-1)d]+\ldots+[2a+(n-1)d]$ | dM1 | For adding; must have $2S$ and be genuine attempt; dependent on 1st M1 |
| $S=\frac{n}{2}[2a+(n-1)d]$ | A1 c.s.o. | Allow first 3 marks for use of $l$ for last term but as given for final mark |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a=149$, $d=-2$; $u_{21}=149+20(-2)=\pounds109$ | M1 A1 | For using $a=149$ and $d=\pm 2$ in $a+(n-1)d$ formula |

### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_n=\frac{n}{2}[2\times149+(n-1)(-2)]$ $(=n(150-n))$ | M1 A1 | For using their $a$, $d$ in $S_n$; A1 any correct expression |
| $S_n=5000 \Rightarrow n^2-150n+5000=0$ | A1 c.s.o. | For putting $S_n=5000$ and simplifying to given expression; no wrong work |

### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(n-100)(n-50)=0$ | M1 | Attempt to solve leading to $n=\ldots$ |
| $n=50$ or $100$ | A2/1/0 | Give A1A0 for 1 correct value; A1A1 for both correct |

### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u_{100}<0$ $\therefore n=100$ not sensible | B1 f.t. | Must mention 100 and state $u_{100}<0$ (or loan paid or equivalent); if giving f.t. then must have $n\geq 76$ |

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Show LaTeX source

9. An arithmetic series has first term $a$ and common difference $d$.
\begin{enumerate}[label=(\alph*)]
\item Prove that the sum of the first $n$ terms of the series is

$$\frac { 1 } { 2 } n [ 2 a + ( n - 1 ) d ] .$$

Sean repays a loan over a period of $n$ months. His monthly repayments form an arithmetic sequence.

He repays $\pounds 149$ in the first month, $\pounds 147$ in the second month, $\pounds 145$ in the third month, and so on. He makes his final repayment in the $n$th month, where $n > 21$.
\item Find the amount Sean repays in the 21st month.

Over the $n$ months, he repays a total of $\pounds 5000$.
\item Form an equation in $n$, and show that your equation may be written as

$$n ^ { 2 } - 150 n + 5000 = 0 .$$
\item Solve the equation in part (c).
\item State, with a reason, which of the solutions to the equation in part (c) is not a sensible solution to the repayment problem.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1 2005 Q9 [13]}}

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