| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Topic | Geometric Sequences and Series |
| Type | Shared terms between AP and GP |
| Difficulty | Standard +0.3 This is a standard sequences problem requiring algebraic manipulation to connect AP and GP terms. Part (i) involves setting up equations from the given conditions and solving simultaneously—routine A-level technique. Part (ii) is a direct application of the GP sum formula. The 7-mark allocation reflects multiple steps rather than conceptual difficulty, and the problem follows a well-established pattern seen in textbooks. |
| 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 |
|---|---|---|
| (i) \(a, a + 8d\) and \(a + 13d\) | M1 | State \(n\)th term of an A.P for at least one term. Must be correct formula. Equate to \(ar\) and \(ar^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{a+13d}{a+8d}\) or \(\frac{a+8d}{a}\) | B1 | State an expression for \(r\), \(d\) or \(r^2\) |
| \(a(a + 13d) = (a + 8d)^2\) | M1 | Equate 2 expressions and make at least one step to solve |
| \(\left(a = -\frac{64d}{3}\right)\) | A1 | Obtain an expression for \(d\) or \(a\) |
| \(d = -\frac{3a}{64}\) | M1 | Substitute their value for \(d\) or \(a\) to find \(r\) |
| \(r = \frac{5}{8}\) | A1 | |
| (ii) \(S = \frac{8a}{3}\) | M1 | Substitute any \(r\) into \(\frac{a}{1-r}\) |
| A1 | [9] | Obtain \(S\) |
(i) $a, a + 8d$ and $a + 13d$ | M1 | State $n$th term of an A.P for at least one term. Must be correct formula. Equate to $ar$ and $ar^2$
$a + 8d = ar$
$a + 13d = ar^2$
$r = \frac{a+13d}{a+8d}$ or $\frac{a+8d}{a}$ | B1 | State an expression for $r$, $d$ or $r^2$
$a(a + 13d) = (a + 8d)^2$ | M1 | Equate 2 expressions and make at least one step to solve
$\left(a = -\frac{64d}{3}\right)$ | A1 | Obtain an expression for $d$ or $a$
$d = -\frac{3a}{64}$ | M1 | Substitute their value for $d$ or $a$ to find $r$
$r = \frac{5}{8}$ | A1 |
(ii) $S = \frac{8a}{3}$ | M1 | Substitute any $r$ into $\frac{a}{1-r}$
| A1 | [9] | Obtain $S$An arithmetic progression has first term $a$ and common difference $d$. The first, ninth and fourteenth terms are, respectively, the first three terms of a geometric progression with common ratio $r$, where $r \neq 1$.
\begin{enumerate}[label=(\roman*)]
\item Find $d$ in terms of $a$ and show that $r = \frac{5}{3}$. [7]
\item Find the sum to infinity of the geometric progression in terms of $a$. [2]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2011 Q11 [9]}}