| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compound growth applications |
| Difficulty | Moderate -0.8 Part (a) is a standard arithmetic progression problem requiring finding common difference and number of terms, then applying the sum formula. Part (b) is a straightforward geometric series application with explicit percentage growth rate and clear time period - both are routine textbook exercises requiring only direct formula application with no problem-solving insight needed. |
| 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 | Marks | Guidance |
| \(a = 32\), \(a + 4d = 22\), \(\rightarrow d = -2.5\) | B1 | |
| \(a + (n-1)d = -28 \rightarrow n = 25\) | B1 | |
| \(S_{25} = \frac{25}{2}(64 - 2.5 \times 24) = 50\) | M1 A1 | M1 for correct formula with \(n = 24\) or \(n = 25\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 2000\), \(r = 1.025\) | B1 | \(r = 1 + 2.5\%\) ok if used correctly in \(S_n\) formula |
| \(S_{10} = 2000\left(\frac{1.025^{10}-1}{1.025-1}\right) = 22400\) or a value which rounds to this | M1 A1 | M1 for correct formula with \(n = 9\) or \(n = 10\) and their \(a\) and \(r\). SR: correct answer only for \(n=10\) B3, for \(n=9\), B1 (£19 900) |
| Total: 3 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 32$, $a + 4d = 22$, $\rightarrow d = -2.5$ | **B1** | |
| $a + (n-1)d = -28 \rightarrow n = 25$ | **B1** | |
| $S_{25} = \frac{25}{2}(64 - 2.5 \times 24) = 50$ | **M1 A1** | **M1** for correct formula with $n = 24$ or $n = 25$ |
| **Total: 4** | | |
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## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 2000$, $r = 1.025$ | **B1** | $r = 1 + 2.5\%$ ok if used correctly in $S_n$ formula |
| $S_{10} = 2000\left(\frac{1.025^{10}-1}{1.025-1}\right) = 22400$ or a value which rounds to this | **M1 A1** | **M1** for correct formula with $n = 9$ or $n = 10$ and their $a$ and $r$. SR: correct answer only for $n=10$ **B3**, for $n=9$, **B1** (£19 900) |
| **Total: 3** | | |4
\begin{enumerate}[label=(\alph*)]
\item An arithmetic progression has a first term of 32, a 5th term of 22 and a last term of - 28 . Find the sum of all the terms in the progression.
\item Each year a school allocates a sum of money for the library. The amount allocated each year increases by $2.5 \%$ of the amount allocated the previous year. In 2005 the school allocated $\$ 2000$. Find the total amount allocated in the years 2005 to 2014 inclusive.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2017 Q4 [7]}}