| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Total over time period |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric sequences requiring students to verify given results and sum a GP. Parts (a) and (b) are 'show that' questions with clear paths, and part (c) involves a standard GP sum formula. The context is accessible and the mathematics is routine for A-level, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Total time for 6 km \(= 24\) minutes \(+ 6\times1.05 + 6\times1.05^2\) minutes | M1 | 3.4 |
| \(= 36.915\) minutes \(=\) 36 minutes 55 seconds | A1* | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Hence time for \(r^{\text{th}}\) km is \(6\times1.05^{r-4}\) | B1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(24 + \displaystyle\sum_{r=5}^{r=20} 6\times1.05^{r-4}\) minutes | M1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| Time for \(5^{\text{th}}\) to \(20^{\text{th}}\) km \(= \dfrac{6.3(1.05^{16}-1)}{1.05-1} = (149.04)\) | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Total time \(= 24 + \dfrac{6.3(1.05^{16}-1)}{1.05-1}\) | A1 | 1.1b |
| Total time \(=\) awrt 173 minutes and 3 seconds | A1 | 1.1b |
# Question 11:
**Part (a):**
Total time for 6 km $= 24$ minutes $+ 6\times1.05 + 6\times1.05^2$ minutes | M1 | 3.4
$= 36.915$ minutes $=$ 36 minutes 55 seconds | A1* | 1.1b
**Total: (2)**
---
**Part (b):**
$5^{\text{th}}$ km is $6\times1.05 = 6\times1.05^1$
$6^{\text{th}}$ km is $6\times1.05\times1.05 = 6\times1.05^2$
$7^{\text{th}}$ km is $6\times1.05^3$
Hence time for $r^{\text{th}}$ km is $6\times1.05^{r-4}$ | B1 | 3.4
**Total: (1)**
---
**Part (c):**
Attempts total time for race:
$24 + \displaystyle\sum_{r=5}^{r=20} 6\times1.05^{r-4}$ minutes | M1 | 3.1a
Uses series formula to find allowable sum:
Time for $5^{\text{th}}$ to $20^{\text{th}}$ km $= \dfrac{6.3(1.05^{16}-1)}{1.05-1} = (149.04)$ | M1 | 3.4
Correct calculation leading to total time:
Total time $= 24 + \dfrac{6.3(1.05^{16}-1)}{1.05-1}$ | A1 | 1.1b
Total time $=$ awrt 173 minutes and 3 seconds | A1 | 1.1b
**Total: (4)**
---\begin{enumerate}
\item A competitor is running a 20 kilometre race.
\end{enumerate}
She runs each of the first 4 kilometres at a steady pace of 6 minutes per kilometre. After the first 4 kilometres, she begins to slow down.
In order to estimate her finishing time, the time that she will take to complete each subsequent kilometre is modelled to be $5 \%$ greater than the time that she took to complete the previous kilometre.
Using the model,\\
(a) show that her time to run the first 6 kilometres is estimated to be 36 minutes 55 seconds,\\
(b) show that her estimated time, in minutes, to run the $r$ th kilometre, for $5 \leqslant r \leqslant 20$, is
$$6 \times 1.05 ^ { r - 4 }$$
(c) estimate the total time, in minutes and seconds, that she will take to complete the race.
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q11 [7]}}