| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Arithmetic/Geometric Series Applications |
| Difficulty | Standard +0.8 This question requires understanding of both arithmetic and geometric series, applying them to a real-world context, and working with infinite series convergence. Part (i) involves calculating sums of 20 terms for both models (requiring careful tracking of up and down distances), while part (ii) requires recognizing and evaluating an infinite geometric series sum. The multi-step nature, need to distinguish between two models, and the infinite series proof elevate this above a standard series question, though the individual techniques are A-level appropriate. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i)(a) \(1.92 + 1.84 + 1.76 + \ldots\) oe | B1 | OR \(a=0.96\), \(d=-0.04\) & ans |
| \(\frac{20}{2}[2\times1.92 + 19\times(-0.08)]\) oe | M1 | doubled/adjusted |
| \(23.2\) cao | A1 | Corr formula used with corr \(d\) & *their* \(a, n\); \(a=1, n=21\rightarrow 12.6\ (25.2)\); \(a=0.96, n=21\rightarrow 11.76\ (23.52)\) |
| [3] | ||
| (i)(b) \(1.92 + 1.92(0.96) + 1.92(0.96)^2 + \ldots\) | B1 | |
| \(\frac{1.92(1 - 0.96^{20})}{1 - 0.96}\) | M1 | OR \(a=0.96\), \(r=0.96\) & ans /doubled/adjusted |
| \(26.8\) cao | A1 | Corr formula used with \(r=0.96\) & *their* \(a, n\); \(a=0.96, n=21\rightarrow13.82\ (27.63)\); \(a=1, n=21\rightarrow14.39\ (28.78)\) |
| [3] | ||
| (ii) \(\frac{1.92}{1-0.96} = 48\) or \(\frac{0.96}{1-0.96} = 24\) & then Double AG | M1A1 | \(a=1\rightarrow25\ (50)\) but must be doubled for M1; \(1.92\frac{(1-0.96^n)}{1-0.96} < 48 \rightarrow 0.96^n > 0\) (www); 'which is true' scores SCB1 |
| [2] |
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)(a)** $1.92 + 1.84 + 1.76 + \ldots$ oe | B1 | OR $a=0.96$, $d=-0.04$ & ans |
| $\frac{20}{2}[2\times1.92 + 19\times(-0.08)]$ oe | M1 | doubled/adjusted |
| $23.2$ cao | A1 | Corr formula used with corr $d$ & *their* $a, n$; $a=1, n=21\rightarrow 12.6\ (25.2)$; $a=0.96, n=21\rightarrow 11.76\ (23.52)$ |
| **[3]** | | |
| **(i)(b)** $1.92 + 1.92(0.96) + 1.92(0.96)^2 + \ldots$ | B1 | |
| $\frac{1.92(1 - 0.96^{20})}{1 - 0.96}$ | M1 | OR $a=0.96$, $r=0.96$ & ans /doubled/adjusted |
| $26.8$ cao | A1 | Corr formula used with $r=0.96$ & *their* $a, n$; $a=0.96, n=21\rightarrow13.82\ (27.63)$; $a=1, n=21\rightarrow14.39\ (28.78)$ |
| **[3]** | | |
| **(ii)** $\frac{1.92}{1-0.96} = 48$ or $\frac{0.96}{1-0.96} = 24$ & then Double **AG** | M1A1 | $a=1\rightarrow25\ (50)$ but must be doubled for M1; $1.92\frac{(1-0.96^n)}{1-0.96} < 48 \rightarrow 0.96^n > 0$ (www); 'which is true' scores SCB1 |
| **[2]** | | |
---6 A ball is such that when it is dropped from a height of 1 metre it bounces vertically from the ground to a height of 0.96 metres. It continues to bounce on the ground and each time the height the ball reaches is reduced. Two different models, $A$ and $B$, describe this.
Model A: The height reached is reduced by 0.04 metres each time the ball bounces.\\
Model B: The height reached is reduced by $4 \%$ each time the ball bounces.\\
(i) Find the total distance travelled vertically (up and down) by the ball from the 1st time it hits the ground until it hits the ground for the 21st time,
\begin{enumerate}[label=(\alph*)]
\item using model $A$,
\item using model $B$.\\
(ii) Show that, under model $B$, even if there is no limit to the number of times the ball bounces, the total vertical distance travelled after the first time it hits the ground cannot exceed 48 metres.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2015 Q6 [8]}}