| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Convergence conditions |
| Difficulty | Moderate -0.8 This question tests standard formula recall and direct substitution. Part (a) requires knowing S_∞ = a/(1-r) and solving a simple equation. Part (b) uses basic AP formulas with straightforward algebra. Both parts are routine textbook exercises with no problem-solving insight required, making them easier than average A-level questions. |
| 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) \(\frac{a}{1-r} = 8a \Rightarrow l(a) = 8(a)(1-r)\) | B1 | |
| \(r = \frac{7}{8}\) oe | B1 | |
| [2] | ||
| (b) \(a + 4d = 197\) | B1 | Or \(2a + 9d = 408\) |
| \(\frac{10}{2}[2a + 9d] = 2040\) | B1 | Attempt to solve simultaneously |
| \(d = 14\) | M1A1 | |
| [4] |
**(a)** $\frac{a}{1-r} = 8a \Rightarrow l(a) = 8(a)(1-r)$ | B1 |
$r = \frac{7}{8}$ oe | B1 |
| | [2]
**(b)** $a + 4d = 197$ | B1 | Or $2a + 9d = 408$
$\frac{10}{2}[2a + 9d] = 2040$ | B1 | Attempt to solve simultaneously
$d = 14$ | M1A1 |
| | [4]5
\begin{enumerate}[label=(\alph*)]
\item In a geometric progression, the sum to infinity is equal to eight times the first term. Find the common ratio.
\item In an arithmetic progression, the fifth term is 197 and the sum of the first ten terms is 2040. Find the common difference.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2013 Q5 [6]}}