| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | New GP from transformation |
| Difficulty | Standard +0.3 Part (a) is a standard arithmetic series problem requiring the sum formula and solving a quadratic inequality. Part (b) requires recognizing that squaring terms of a GP creates a new GP with first term a² and common ratio r², then applying the sum to infinity formula—a moderately routine application of GP properties with one conceptual step beyond basic recall. |
| 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 |
| \((S_n =)\,\frac{n}{2}[32+(n-1)8]\) and \(20000\) | M1 | M1 correct formula used with \(d\) from \(16+d=24\) |
| A1 | A1 for correct expression linked to \(20000\) | |
| \(\rightarrow n^2 + 3n - 5000\ (<,=,>0)\) | DM1 | Simplification to a three term quadratic |
| \(\rightarrow (n=69.2)\rightarrow 70\) terms needed | A1 | Condone use of \(20001\) throughout. Correct answer from trial and improvement gets 4/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a=6,\ \frac{a}{1-r}=18 \rightarrow r=\frac{2}{3}\) | M1A1 | Correct \(S_\infty\) formula used to find \(r\) |
| New progression \(a=36,\ r=\frac{4}{9}\) oe | M1 | Obtain new values for \(a\) and \(r\) by any valid method |
| New \(S_\infty = \frac{36}{1-\frac{4}{9}} \rightarrow 64.8\) or \(\frac{324}{5}\) oe | A1 | Be aware that \(r=-\frac{2}{3}\) leads to \(64.8\) but can only score M marks |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(S_n =)\,\frac{n}{2}[32+(n-1)8]$ and $20000$ | M1 | **M1** correct formula used with $d$ from $16+d=24$ |
| | A1 | **A1** for correct expression linked to $20000$ |
| $\rightarrow n^2 + 3n - 5000\ (<,=,>0)$ | DM1 | Simplification to a three term quadratic |
| $\rightarrow (n=69.2)\rightarrow 70$ terms needed | A1 | Condone use of $20001$ throughout. Correct answer from trial and improvement gets 4/4 |
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## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a=6,\ \frac{a}{1-r}=18 \rightarrow r=\frac{2}{3}$ | M1A1 | Correct $S_\infty$ formula used to find $r$ |
| New progression $a=36,\ r=\frac{4}{9}$ oe | M1 | Obtain new values for $a$ and $r$ by any valid method |
| New $S_\infty = \frac{36}{1-\frac{4}{9}} \rightarrow 64.8$ or $\frac{324}{5}$ oe | A1 | Be aware that $r=-\frac{2}{3}$ leads to $64.8$ but can only score M marks |
---7
\begin{enumerate}[label=(\alph*)]
\item The first two terms of an arithmetic progression are 16 and 24. Find the least number of terms of the progression which must be taken for their sum to exceed 20000.
\item A geometric progression has a first term of 6 and a sum to infinity of 18. A new geometric progression is formed by squaring each of the terms of the original progression. Find the sum to infinity of the new progression.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2017 Q7 [8]}}