| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Relationship between two GPs |
| Difficulty | Standard +0.3 Part (a) is a standard arithmetic progression problem requiring routine manipulation of sum formulas. Part (b) involves setting up two simultaneous equations from sum-to-infinity formulas (a/(1-r) = 6 and 2a/(1-r²) = 7), then solving—this requires algebraic manipulation but follows a predictable pattern. The question is slightly above average due to the two-part structure and the need to handle r² in part (b), but both parts are textbook-standard exercises requiring no novel insight. |
| 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 |
|---|---|---|
| (a) \(\frac{10}{2}(2a + 9d) = 400\) oe | B1 | \(\rightarrow 2a + 9d = 80\) |
| \(\frac{20}{2}(2a + 19d) = 1400\) OR | ||
| \(\frac{10}{2}[2(a+10d) + 9d] = 1000\) | B1 | \(\rightarrow 2a + 19d = 140\) or \(2a + 29d = 200\) |
| \(d = 6\) \(a = 13\) | M1A1A1 | Solve sim. eqns both from \(S_n\) formulae |
| (b) \(\frac{a}{1-r} = 6\) | B1B1 | |
| \(\frac{2a}{1-r^2} = 7\) | ||
| \(\frac{12(1-r)}{1-r^2} = 7\) or \(\frac{1-r^2}{1-r} = \frac{12}{7}\) | M1 | Substitute or divide |
| \(r = \frac{5}{7}\) or 0.714 | A1 | |
| \(a = \frac{12}{7}\) or 1.71(4) | A1 | Ignore any other solns for \(r\) and \(a\) |
(a) $\frac{10}{2}(2a + 9d) = 400$ oe | B1 | $\rightarrow 2a + 9d = 80$
$\frac{20}{2}(2a + 19d) = 1400$ OR | |
$\frac{10}{2}[2(a+10d) + 9d] = 1000$ | B1 | $\rightarrow 2a + 19d = 140$ or $2a + 29d = 200$
$d = 6$ $a = 13$ | M1A1A1 | Solve sim. eqns both from $S_n$ formulae | [5]
(b) $\frac{a}{1-r} = 6$ | B1B1 |
$\frac{2a}{1-r^2} = 7$ | |
$\frac{12(1-r)}{1-r^2} = 7$ or $\frac{1-r^2}{1-r} = \frac{12}{7}$ | M1 | Substitute or divide
$r = \frac{5}{7}$ or 0.714 | A1 |
$a = \frac{12}{7}$ or 1.71(4) | A1 | Ignore any other solns for $r$ and $a$ | [5]9
\begin{enumerate}[label=(\alph*)]
\item In an arithmetic progression the sum of the first ten terms is 400 and the sum of the next ten terms is 1000 . Find the common difference and the first term.
\item A geometric progression has first term $a$, common ratio $r$ and sum to infinity 6. A second geometric progression has first term $2 a$, common ratio $r ^ { 2 }$ and sum to infinity 7 . Find the values of $a$ and $r$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2013 Q9 [10]}}