| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sequence defined by formula |
| Difficulty | Easy -1.2 Part (a) is a straightforward compound interest calculation requiring a single application of the formula (10000 × 1.05^10). Part (b) involves finding the first term and common difference from a given sum formula, which is a standard textbook exercise requiring knowledge that S_n - S_{n-1} = a_n and using S_1 = a_1. Both parts are routine applications of well-practiced techniques with no problem-solving insight required. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses \(r = (1.05 \text{ or } 105\%)^{9, 10 \text{ or } 11}\) | B1 | Used to multiply repeatedly or in any GP formula. |
| New value \(= 10000 \times 1.05^{10} = (\\))16\,300$ | B1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: \(n=1 \to 5\), \(a=5\) | (B1) | Uses \(n=1\) to find \(a\) |
| \(n=2 \to 13\) | B1 | Correct \(S_n\) for any other value of \(n\) (e.g. \(n=2\)) |
| \(a+(a+d)=13 \to d=3\) | M1 A1 | Correct method leading to \(d=\) |
| OR: \(\left(\frac{n}{2}\right)(2a+(n-1)d) = \left(\frac{n}{2}\right)(3n+7)\) | \(\left(\frac{n}{2}\right)\) maybe be ignored | |
| \(\therefore dn+2a-d=3n+7 \to dn=3n \to d=3\) | (*M1A1) | Method mark awarded for equating terms in \(n\) from correct \(S_n\) formula. |
| \(2a - (\text{their } 3) = 7\), \(a=5\) | DM1 A1 | |
| 4 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses $r = (1.05 \text{ or } 105\%)^{9, 10 \text{ or } 11}$ | **B1** | Used to multiply repeatedly or in any GP formula. |
| New value $= 10000 \times 1.05^{10} = (\$)16\,300$ | **B1** | |
| | **2** | |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** $n=1 \to 5$, $a=5$ | **(B1)** | Uses $n=1$ to find $a$ |
| $n=2 \to 13$ | **B1** | Correct $S_n$ for any other value of $n$ (e.g. $n=2$) |
| $a+(a+d)=13 \to d=3$ | **M1 A1** | Correct method leading to $d=$ |
| **OR:** $\left(\frac{n}{2}\right)(2a+(n-1)d) = \left(\frac{n}{2}\right)(3n+7)$ | | $\left(\frac{n}{2}\right)$ maybe be ignored |
| $\therefore dn+2a-d=3n+7 \to dn=3n \to d=3$ | **(*M1A1)** | Method mark awarded for equating terms in $n$ from correct $S_n$ formula. |
| $2a - (\text{their } 3) = 7$, $a=5$ | **DM1 A1** | |
| | **4** | |3
\begin{enumerate}[label=(\alph*)]
\item Each year, the value of a certain rare stamp increases by $5 \%$ of its value at the beginning of the year. A collector bought the stamp for $\$ 10000$ at the beginning of 2005. Find its value at the beginning of 2015 correct to the nearest $\$ 100$.
\item The sum of the first $n$ terms of an arithmetic progression is $\frac { 1 } { 2 } n ( 3 n + 7 )$. Find the 1 st term and the common difference of the progression.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2017 Q3 [6]}}