| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Real-world AP: find term or total |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on arithmetic and geometric sequences with clear structure. Parts (a)-(c) involve direct application of standard formulas (nth term, sum of AP, GP ratio), while part (d) requires solving an inequality but with guided setup. The real-world context doesn't add conceptual difficulty, and all steps follow predictable patterns slightly easier than a typical A-level question. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_{20} = 100 + 19(-2) = 62\) | B1* | Correct method shown; accept seeing \(100+19(-2)=62\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{20} = \frac{1}{2}(20)\{2\times100+19(-2)\}\) or \(S_{20}=\frac{1}{2}(20)\{100+62\}\) | M1 | Applies correct AP sum formula with \(n=20\), \(a=100\), \(d=-2\) or \(l=62\) |
| \(= 1620\text{ (mm)}\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(62\times r^2 = 60 \Rightarrow r^2 = \ldots\) | M1 | Correct strategy to find \(r\) |
| \(r^2 = \frac{60}{62} \Rightarrow r = \sqrt{\frac{60}{62}} = 0.983738\ldots\) | A1 | \(r\) = awrt 0.984; accept exact answer \(\sqrt{\frac{30}{31}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Total distance from GS hits \(= \frac{62\times0.983\ldots(1-0.983\ldots^n)}{1-0.983\ldots}\) | M1 | Applies correct GS summation with their \(r\), \(a=62\) or \(62\times\) their \(r\), and \(n\) or \(n-1\) or \(n-19\) or \(n-20\) |
| \(1620 + \frac{62\times0.983\ldots(1-0.983\ldots^n)}{1-0.983\ldots} > 3000\) | M1 | Correct equation set up with their \(r\) and suitable \(a\) |
| \(0.983\ldots^n < 0.63207\ldots \Rightarrow n = \frac{\log(0.63207\ldots)}{\log(0.983\ldots)}\) | M1 | Fully correct processing to find \(n\) from equation of suitable form |
| \(n = 27.98\ldots \Rightarrow N = 20+28 = 48\) | A1cso | \(N=48\) only |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_{20} = 100 + 19(-2) = 62$ | B1* | Correct method shown; accept seeing $100+19(-2)=62$ |
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{20} = \frac{1}{2}(20)\{2\times100+19(-2)\}$ or $S_{20}=\frac{1}{2}(20)\{100+62\}$ | M1 | Applies correct AP sum formula with $n=20$, $a=100$, $d=-2$ or $l=62$ |
| $= 1620\text{ (mm)}$ | A1 | Correct value |
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $62\times r^2 = 60 \Rightarrow r^2 = \ldots$ | M1 | Correct strategy to find $r$ |
| $r^2 = \frac{60}{62} \Rightarrow r = \sqrt{\frac{60}{62}} = 0.983738\ldots$ | A1 | $r$ = awrt 0.984; accept exact answer $\sqrt{\frac{30}{31}}$ |
## Question 8(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Total distance from GS hits $= \frac{62\times0.983\ldots(1-0.983\ldots^n)}{1-0.983\ldots}$ | M1 | Applies correct GS summation with their $r$, $a=62$ or $62\times$ their $r$, and $n$ or $n-1$ or $n-19$ or $n-20$ |
| $1620 + \frac{62\times0.983\ldots(1-0.983\ldots^n)}{1-0.983\ldots} > 3000$ | M1 | Correct equation set up with their $r$ and suitable $a$ |
| $0.983\ldots^n < 0.63207\ldots \Rightarrow n = \frac{\log(0.63207\ldots)}{\log(0.983\ldots)}$ | M1 | Fully correct processing to find $n$ from equation of suitable form |
| $n = 27.98\ldots \Rightarrow N = 20+28 = 48$ | A1cso | $N=48$ only |8. A metal post is repeatedly hit in order to drive it into the ground.
Given that
\begin{itemize}
\item on the 1st hit, the post is driven 100 mm into the ground
\item on the 2nd hit, the post is driven an additional 98 mm into the ground
\item on the 3rd hit, the post is driven an additional 96 mm into the ground
\item the additional distances the post travels on each subsequent hit form an arithmetic sequence
\begin{enumerate}[label=(\alph*)]
\item show that the post is driven an additional 62 mm into the ground with the 20th hit.
\item Find the total distance that the post has been driven into the ground after 20 hits.
\end{itemize}
Given that for each subsequent hit after the 20th hit
\begin{itemize}
\item the additional distances the post travels form a geometric sequence with common ratio $r$
\item on the 22 nd hit, the post is driven an additional 60 mm into the ground
\item find the value of $r$, giving your answer to 3 decimal places.
\end{itemize}
After a total of $N$ hits, the post will have been driven more than 3 m into the ground.
\item Find, showing all steps in your working, the smallest possible value of $N$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2022 Q8 [9]}}