Edexcel C2 — Question 8 13 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeProve sum formula
DifficultyModerate -0.8 Part (a) is a standard bookwork proof of the arithmetic series formula that appears in every C2 textbook. Parts (b)-(d) are routine applications requiring substitution into formulas and basic algebraic manipulation. The multi-part structure adds length but not conceptual difficulty—each step is straightforward with no problem-solving insight required.
Spec1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum

8. (a) An arithmetic series has first term \(a\) and common difference \(d\). Prove that the sum of the first \(n\) terms of the series is \(\frac { 1 } { 2 } n [ 2 a + ( n - 1 ) d ]\). A company made a profit of \(\pounds 54000\) in the year 2001. A model for future performance assumes that yearly profits will increase in an arithmetic sequence with common difference \(\pounds d\). This model predicts total profits of \(\pounds 619200\) for the 9 years 2001 to 2009 inclusive.
(b) Find the value of \(d\). Using your value of \(d\),
(c) find the predicted profit for the year 2011. An alternative model assumes that the company's yearly profits will increase in a geometric sequence with common ratio 1.06 . Using this alternative model and again taking the profit in 2001 to be \(\pounds 54000\),
(d) find the predicted profit for the year 2011.

Question 8:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(S = a + (a+d) + (a+2d) + \ldots + [a+(n-1)d]\)B1
\(S = [a+(n-1)d] + [a+(n-2)d] + \ldots + a\)M1
Add: \(2S = n[2a+(n-1)d] \Rightarrow S = \frac{1}{2}n[2a+(n-1)d]\)M1 A1 (4)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(a = 54000\) and \(n = 9\)B1
\(619200 = \frac{1}{2}\times 9 \times (2\times 54000 + 8d)\)M1 A1ft
\(d = 3700\)A1 (4)
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(a + (n-1)d = a + 10d = 54000 + 10d = £91000\)M1 A1 (2)
Part (d)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(ar^{n-1} = 54000 \times 1.06^{10}\) (ft their \(n\))M1 A1ft
\(= £96700\) (or £97000)A1 (3)
## Question 8:

### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S = a + (a+d) + (a+2d) + \ldots + [a+(n-1)d]$ | B1 | |
| $S = [a+(n-1)d] + [a+(n-2)d] + \ldots + a$ | M1 | |
| Add: $2S = n[2a+(n-1)d] \Rightarrow S = \frac{1}{2}n[2a+(n-1)d]$ | M1 A1 | (4) |

### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = 54000$ and $n = 9$ | B1 | |
| $619200 = \frac{1}{2}\times 9 \times (2\times 54000 + 8d)$ | M1 A1ft | |
| $d = 3700$ | A1 | (4) |

### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a + (n-1)d = a + 10d = 54000 + 10d = £91000$ | M1 A1 | (2) |

### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $ar^{n-1} = 54000 \times 1.06^{10}$ (ft their $n$) | M1 A1ft | |
| $= £96700$ (or £97000) | A1 | (3) |
8. (a) An arithmetic series has first term $a$ and common difference $d$. Prove that the sum of the first $n$ terms of the series is $\frac { 1 } { 2 } n [ 2 a + ( n - 1 ) d ]$.

A company made a profit of $\pounds 54000$ in the year 2001. A model for future performance assumes that yearly profits will increase in an arithmetic sequence with common difference $\pounds d$. This model predicts total profits of $\pounds 619200$ for the 9 years 2001 to 2009 inclusive.\\
(b) Find the value of $d$.

Using your value of $d$,\\
(c) find the predicted profit for the year 2011.

An alternative model assumes that the company's yearly profits will increase in a geometric sequence with common ratio 1.06 . Using this alternative model and again taking the profit in 2001 to be $\pounds 54000$,\\
(d) find the predicted profit for the year 2011.

\hfill \mbox{\textit{Edexcel C2  Q8 [13]}}