| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compound growth applications |
| Difficulty | Moderate -0.8 This is a straightforward application of geometric sequences requiring only direct formula substitution. Part (a) is simple verification (120000 × 1.05³), part (b) requires solving 120000 × 1.05^n > 200000 using logarithms (standard C2 technique), and part (c) applies the geometric series sum formula. All steps are routine with no problem-solving insight needed, making this easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(120000 \times (1.05)^3 = 138915\) | B1 | Or \(120000 \times 1.05 \times 1.05 \times 1.05 = 138915\); Or \(120000, 126000, 132300, 138915\); Or \(a = 120000\) and \(a \times (1.05)^3 = 138915\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(120000 \times (1.05)^{n-1} > 200000\) | M1 | Allow \(n\) or \(n-1\) and "\(>\)", "\(<\)", or "\(=\)" etc. |
| \(\log 1.05^{n-1} > \log\left(\dfrac{5}{3}\right)\) | M1 | Takes logs correctly. Allow \(n\) or \(n-1\) and "\(>\)", "\(<\)", or "\(=\)" etc. |
| \((n-1) > \dfrac{\log\left(\frac{5}{3}\right)}{\log 1.05}\) or equivalent, e.g. \(\left(n > \dfrac{\log\left(\frac{7}{4}\right)}{\log 1.05}\right)\) | A1 | Allow \(n\) or \(n-1\) and "\(>\)", "\(<\)", or "\(=\)" etc. Allow \(1.\dot{6}\) or awrt \(1.67\) for \(5/3\) |
| \(2024\) | M1A1 | M1: Identifies a calendar year using their value of \(n\) or \(n-1\). A1: 2024 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{a(1-r^n)}{1-r} = \dfrac{120000(1-1.05^{11})}{1-1.05}\) | M1A1 | M1: Correct sum formula with \(n = 10, 11\) or \(12\). A1: Correct numerical expression with \(n = 11\) |
| \(1704814\) | A1 | Cao (Allow 1704814.00) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(U_{10} = 186159.39,\ U_{11} = 195467.36,\ U_{12} = 205240.72\). Attempt to find at least the \(10^{\text{th}}\) or \(11^{\text{th}}\) or \(12^{\text{th}}\) terms correctly using common ratio of \(1.05\) (all terms need not be listed) | M1 | |
| Forms the geometric progression correctly to reach a term \(> 200000\) | M1 | |
| Obtains an "\(11^{\text{th}}\)" term of awrt \(195500\) and a "\(12^{\text{th}}\)" term of awrt \(205200\) | A1 | |
| Uses their number of terms to identify a calendar year | M1 | |
| \(2024\) | A1 |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $120000 \times (1.05)^3 = 138915$ | B1 | Or $120000 \times 1.05 \times 1.05 \times 1.05 = 138915$; Or $120000, 126000, 132300, 138915$; Or $a = 120000$ and $a \times (1.05)^3 = 138915$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $120000 \times (1.05)^{n-1} > 200000$ | M1 | Allow $n$ or $n-1$ and "$>$", "$<$", or "$=$" etc. |
| $\log 1.05^{n-1} > \log\left(\dfrac{5}{3}\right)$ | M1 | Takes logs correctly. Allow $n$ or $n-1$ and "$>$", "$<$", or "$=$" etc. |
| $(n-1) > \dfrac{\log\left(\frac{5}{3}\right)}{\log 1.05}$ or equivalent, e.g. $\left(n > \dfrac{\log\left(\frac{7}{4}\right)}{\log 1.05}\right)$ | A1 | Allow $n$ or $n-1$ and "$>$", "$<$", or "$=$" etc. Allow $1.\dot{6}$ or awrt $1.67$ for $5/3$ |
| $2024$ | M1A1 | M1: Identifies a calendar year using their value of $n$ or $n-1$. A1: 2024 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{a(1-r^n)}{1-r} = \dfrac{120000(1-1.05^{11})}{1-1.05}$ | M1A1 | M1: Correct sum formula with $n = 10, 11$ or $12$. A1: Correct numerical expression with $n = 11$ |
| $1704814$ | A1 | Cao (Allow 1704814.00) |
**Listing/trial method for (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $U_{10} = 186159.39,\ U_{11} = 195467.36,\ U_{12} = 205240.72$. Attempt to find at least the $10^{\text{th}}$ or $11^{\text{th}}$ or $12^{\text{th}}$ terms correctly using common ratio of $1.05$ (all terms need not be listed) | M1 | |
| Forms the geometric progression correctly to reach a term $> 200000$ | M1 | |
| Obtains an "$11^{\text{th}}$" term of awrt $195500$ and a "$12^{\text{th}}$" term of awrt $205200$ | A1 | |
| Uses their number of terms to identify a calendar year | M1 | |
| $2024$ | A1 | |
---3. A company predicts a yearly profit of $\pounds 120000$ in the year 2013 . The company predicts that the yearly profit will rise each year by $5 \%$. The predicted yearly profit forms a geometric sequence with common ratio 1.05
\begin{enumerate}[label=(\alph*)]
\item Show that the predicted profit in the year 2016 is $\pounds 138915$
\item Find the first year in which the yearly predicted profit exceeds $\pounds 200000$
\item Find the total predicted profit for the years 2013 to 2023 inclusive, giving your answer to the nearest pound.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q3 [9]}}