| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compound growth applications |
| Difficulty | Moderate -0.8 This is a straightforward application of arithmetic sequences (not geometric sequences despite part b) requiring basic formula substitution. Part (a) uses nth term and sum formulas with simple arithmetic; part (b) involves standard GP equations with routine algebraic manipulation. All techniques are textbook exercises with no novel problem-solving required, making it easier than average. |
| 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 |
|---|---|---|
| \(200 + (15-1)(+/-5)\) | M1 | Use of \(n\)th term with \(a = 200\), \(n = 14\) or \(15\) and \(d = +/-5\) |
| \(= 130\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{n}{2}[400 + (n-1)(+/-5)] = 3050\) | M1 | Use of \(S_n\), \(a=200\) and \(d = +/-5\) |
| \(\rightarrow 5n^2 - 405n + 6100 = 0\) | A1 | |
| \(\rightarrow 20\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(ar^2, ar^5 \rightarrow r = \frac{1}{2}\) | M1 A1 | Both terms correct |
| \(\frac{63}{2} = \frac{a(1 - \frac{1}{2}^6)}{\frac{1}{2}} \rightarrow a = 16\) | M1 A1 | Use of \(S_n = 31.5\) with a numeric \(r\) [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Sum to infinity \(= \frac{16}{\frac{1}{2}} = 32\) | B1✓ | ✓ for their \(a\) and \(r\) with \( |
# Question 8:
## Part (a)(i):
$200 + (15-1)(+/-5)$ | M1 | Use of $n$th term with $a = 200$, $n = 14$ or $15$ and $d = +/-5$
$= 130$ | A1 | [2]
## Part (a)(ii):
$\frac{n}{2}[400 + (n-1)(+/-5)] = 3050$ | M1 | Use of $S_n$, $a=200$ and $d = +/-5$
$\rightarrow 5n^2 - 405n + 6100 = 0$ | A1 |
$\rightarrow 20$ | A1 | [3]
## Part (b)(i):
$ar^2, ar^5 \rightarrow r = \frac{1}{2}$ | M1 A1 | Both terms correct
$\frac{63}{2} = \frac{a(1 - \frac{1}{2}^6)}{\frac{1}{2}} \rightarrow a = 16$ | M1 A1 | Use of $S_n = 31.5$ with a numeric $r$ [4]
## Part (b)(ii):
Sum to infinity $= \frac{16}{\frac{1}{2}} = 32$ | B1✓ | ✓ for their $a$ and $r$ with $|r| < 1$ [1]
---8
\begin{enumerate}[label=(\alph*)]
\item A cyclist completes a long-distance charity event across Africa. The total distance is 3050 km . He starts the event on May 1st and cycles 200 km on that day. On each subsequent day he reduces the distance cycled by 5 km .
\begin{enumerate}[label=(\roman*)]
\item How far will he travel on May 15th?
\item On what date will he finish the event?
\end{enumerate}\item A geometric progression is such that the third term is 8 times the sixth term, and the sum of the first six terms is $31 \frac { 1 } { 2 }$. Find
\begin{enumerate}[label=(\roman*)]
\item the first term of the progression,
\item the sum to infinity of the progression.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2016 Q8 [10]}}