| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compare two growth models |
| Difficulty | Standard +0.3 Part (a) is a standard geometric progression problem requiring finding the common ratio from consecutive terms and applying the sum to infinity formula. Part (b) involves straightforward application of arithmetic and geometric series formulas to real-world contexts with clear parameters. Both parts are routine A-level exercises with no novel problem-solving required, though the multi-part structure and series summation calculations place it slightly above average difficulty. |
| 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|<11.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(ar^2 = 48\), \(ar^3 = 32\), \(r = \frac{2}{3}\) or \(a = 108\) | M1 | Solution of the 2 equations to give \(r\) (or \(a\)). A1 (both) |
| \(r = \frac{2}{3}\) and \(a = 108\) | A1 | |
| \(S_\infty = \frac{108}{\frac{1}{3}} = 324\) | A1 | FT Needs correct formula and \(r\) between \(-1\) and \(1\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Scheme A: \(a = 2.50\), \(d = 0.16\); \(S_n = 12(5 + 23 \times 0.16)\) | M1 | Correct use of either AP \(S_n\) formula |
| \(S_n = 104\) tonnes | A1 | |
| Scheme B: \(a = 2.50\), \(r = 1.06\) | B1 | Correct value of \(r\) used in GP |
| \(= \frac{2.5(1.06^{24}-1)}{1.06-1}\) | M1 | Correct use of either \(S_n\) formula |
| \(S_n = 127\) tonnes | A1 | |
| Total: 5 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $ar^2 = 48$, $ar^3 = 32$, $r = \frac{2}{3}$ **or** $a = 108$ | M1 | Solution of the 2 equations to give $r$ (or $a$). A1 (both) |
| $r = \frac{2}{3}$ **and** $a = 108$ | A1 | |
| $S_\infty = \frac{108}{\frac{1}{3}} = 324$ | A1 | **FT** Needs correct formula and $r$ between $-1$ and $1$ |
| **Total: 3** | | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Scheme A: $a = 2.50$, $d = 0.16$; $S_n = 12(5 + 23 \times 0.16)$ | M1 | Correct use of either AP $S_n$ formula |
| $S_n = 104$ tonnes | A1 | |
| Scheme B: $a = 2.50$, $r = 1.06$ | B1 | Correct value of $r$ used in GP |
| $= \frac{2.5(1.06^{24}-1)}{1.06-1}$ | M1 | Correct use of either $S_n$ formula |
| $S_n = 127$ tonnes | A1 | |
| **Total: 5** | | |8
\begin{enumerate}[label=(\alph*)]
\item The third and fourth terms of a geometric progression are 48 and 32 respectively. Find the sum to infinity of the progression.
\item Two schemes are proposed for increasing the amount of household waste that is recycled each week.
Scheme $A$ is to increase the amount of waste recycled each month by 0.16 tonnes.\\
Scheme $B$ is to increase the amount of waste recycled each month by $6 \%$ of the amount recycled in the previous month.\\
The proposal is to operate the scheme for a period of 24 months. The amount recycled in the first month is 2.5 tonnes.
For each scheme, find the total amount of waste that would be recycled over the 24 -month period.
Scheme $A$\\
Scheme $B$ $\_\_\_\_$
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2019 Q8 [8]}}