| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Mixed arithmetic and geometric |
| Difficulty | Moderate -0.3 Part (a) is a standard arithmetic progression problem requiring setup of simultaneous equations (using a+3d and sum formula) and solving for n - routine but multi-step. Part (b) involves finding the common ratio from a given condition (r²=4, so r=±2) then applying the GP sum formula - straightforward application of formulas with minor algebraic manipulation. Both parts are typical textbook exercises requiring no novel insight, slightly easier than average due to being Pure Maths 1 level with standard techniques. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(57 = 2(24 + 3d) \rightarrow d = 1.5\). \(48 = 12 + (n-1)1.5 \rightarrow n = 25\) | M1 A1 M1 A1 | Use of correct \(S_n\) formula. Use of correct \(T_n\) formula |
| [4] | ||
| (b) \(ar^2 = 4a\), \(r = \pm 2\). \(\frac{a(r^6-1)}{r-1} = ka\) → \(k = 63\) or \(k = -21\) | B1 B1 B1 B1 | (allow for \(r = 2\)) |
| [4] |
(a) $57 = 2(24 + 3d) \rightarrow d = 1.5$. $48 = 12 + (n-1)1.5 \rightarrow n = 25$ | M1 A1 M1 A1 | Use of correct $S_n$ formula. Use of correct $T_n$ formula
[4] |
(b) $ar^2 = 4a$, $r = \pm 2$. $\frac{a(r^6-1)}{r-1} = ka$ → $k = 63$ or $k = -21$ | B1 B1 B1 B1 | (allow for $r = 2$)
[4] |
---10
\begin{enumerate}[label=(\alph*)]
\item The first and last terms of an arithmetic progression are 12 and 48 respectively. The sum of the first four terms is 57. Find the number of terms in the progression.
\item The third term of a geometric progression is four times the first term. The sum of the first six terms is $k$ times the first term. Find the possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2013 Q10 [8]}}