Edexcel C1 — Question 9

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeProve sum formula
DifficultyModerate -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 algebraic manipulation. The quadratic in part (d) factors easily, and part (e) requires only basic contextual reasoning. No novel problem-solving or insight required—purely procedural application of a standard topic.
Spec1.02f Solve quadratic equations: including in a function of unknown1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum

9. An arithmetic series has first term \(a\) and common difference \(d\).
  1. 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 \(n > 21\).
  2. Find the amount Sean repays in the 21st month. Over the \(n\) months, he repays a total of \(\pounds 5000\).
  3. Form an equation in \(n\), and show that your equation may be written as $$n ^ { 2 } - 150 n + 5000 = 0$$
  4. Solve the equation in part (c).
  5. State, with a reason, which of the solutions to the equation in part (c) is not a sensible solution to the repayment problem.

Question 9:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(2x+3y=26 \Rightarrow 3y=26-2x\), attempt to find \(m\) from \(y=mx+c\)M1 Complete method for finding gradient; may be implied by later correct work
Gradient \(= -\frac{2}{3}\)A1 Condone \(-\frac{2}{3}x\); ignore errors in constant term
Gradient of perpendicular \(= \frac{-1}{\text{their gradient}} = \frac{3}{2}\)M1 Uses \(m_1 \times m_2 = -1\)
Line through \((0,0)\): \(y = \frac{3}{2}x\)A1 Allow \(2y=3x\), \(y-3x/2=0\), \(2y-3x=0\) etc.
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Solve \(y=\frac{3}{2}x\) with \(2x+3y=26\) to form equation in \(x\) or \(y\)M1 Eliminates variable between the two equations
Solve to obtain \(x=\) or \(y=\)dM1 Depends on previous M; attempts to solve
\(x=4\) or \(y=6\)A1
\(B = \left(0, \frac{26}{3}\right)\) used or statedB1 \(y\)-coordinate of \(B\) is \(\frac{26}{3}\); must be used or stated in (b)
Area \(= \frac{1}{2} \times 4 \times \frac{26}{3}\) (Method 1: \(\frac{1}{2}\times OB \times x\)-coordinate of \(C\))dM1 Depends on previous M; complete method to find area of triangle \(OBC\)
\(= \frac{52}{3}\)A1 cao with integer numerator and denominator; also accept \(\frac{104}{6}\) or \(\frac{1352}{78}\)
# Question 9:

## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x+3y=26 \Rightarrow 3y=26-2x$, attempt to find $m$ from $y=mx+c$ | M1 | Complete method for finding gradient; may be implied by later correct work |
| Gradient $= -\frac{2}{3}$ | A1 | Condone $-\frac{2}{3}x$; ignore errors in constant term |
| Gradient of perpendicular $= \frac{-1}{\text{their gradient}} = \frac{3}{2}$ | M1 | Uses $m_1 \times m_2 = -1$ |
| Line through $(0,0)$: $y = \frac{3}{2}x$ | A1 | Allow $2y=3x$, $y-3x/2=0$, $2y-3x=0$ etc. |

## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solve $y=\frac{3}{2}x$ with $2x+3y=26$ to form equation in $x$ or $y$ | M1 | Eliminates variable between the two equations |
| Solve to obtain $x=$ or $y=$ | dM1 | Depends on previous M; attempts to solve |
| $x=4$ or $y=6$ | A1 | |
| $B = \left(0, \frac{26}{3}\right)$ used or stated | B1 | $y$-coordinate of $B$ is $\frac{26}{3}$; must be used or stated in (b) |
| Area $= \frac{1}{2} \times 4 \times \frac{26}{3}$ (Method 1: $\frac{1}{2}\times OB \times x$-coordinate of $C$) | dM1 | Depends on previous M; complete method to find area of triangle $OBC$ |
| $= \frac{52}{3}$ | A1 | cao with integer numerator and denominator; also accept $\frac{104}{6}$ or $\frac{1352}{78}$ |
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  Q9}}