| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Real-world AP: find n satisfying a condition |
| Difficulty | Moderate -0.8 Part (a) is a straightforward application of arithmetic series formula (S_n = n/2[2a + (n-1)d]) requiring one equation to solve for d, then finding the third term. Part (b) is direct recall of geometric series sum to infinity formula with r = 2/3. Both parts are routine textbook exercises with no problem-solving insight required, making this easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(3^{\text{rd}}\) term \(= a+2d = \\)61.50\( | M1, A1, A1, A1 [3] | Correct formula (M0 if \)n\(th term used). Co. Use of \)a+2d\( with his \)d\(. \)61.5$ ok. |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_\infty = \frac{a}{1-r} = 18\) | M1, M1, A1 [3] | \(a\), ar correct, and \(r\) evaluated. Correct formula used, but needs \(r < 1\) for M mark. |
**(a)** $a=60$, $n=48$, $S_n=3726$
$S_n$ formula used
$\rightarrow d = \$0.75$
$3^{\text{rd}}$ term $= a+2d = \$61.50$ | M1, A1, A1, A1 [3] | Correct formula (M0 if $n$th term used). Co. Use of $a+2d$ with his $d$. $61.5$ ok.
**(b)** $a=6$ ar $=4$ $\therefore r=\frac{2}{3}$
$S_\infty = \frac{a}{1-r} = 18$ | M1, M1, A1 [3] | $a$, ar correct, and $r$ evaluated. Correct formula used, but needs $r < 1$ for M mark.3
\begin{enumerate}[label=(\alph*)]
\item A debt of $\$ 3726$ is repaid by weekly payments which are in arithmetic progression. The first payment is $\$ 60$ and the debt is fully repaid after 48 weeks. Find the third payment.
\item Find the sum to infinity of the geometric progression whose first term is 6 and whose second term is 4 .
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2003 Q3 [6]}}