| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find first term from conditions |
| Difficulty | Moderate -0.3 Part (a) requires setting up two equations (ar=12, a/(1-r)=54) and solving a quadratic, which is standard GP work. Part (b) involves deriving a sum formula for an arithmetic progression and solving simultaneous equations. Both parts are routine applications of formulas with straightforward algebra, making this slightly easier than average A-level difficulty. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(ar = 12\) and \(\frac{a}{1-r} = 54\) | B1 B1 | CAO, OE CAO, OE |
| Eliminates \(a\) or \(r \rightarrow 9r^2 - 9r + 2 = 0\) or \(a^2 - 54a + 648 = 0\) | M1 | Elimination leading to a 3-term quadratic in \(a\) or \(r\) |
| \(r = \frac{2}{3}\) or \(\frac{1}{3}\) hence \(a \rightarrow a = 18\) or \(36\) | A1 | Needs both values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| First term \(= p + q\). Difference \(= q\) or last term \(= p + qn\) | B1 | Need first term and last term or common difference |
| \(S_n = \frac{n}{2}(2(p+q)+(n-1)q)\) or \(\frac{n}{2}(2p+q+nq)\) | M1A1 | Use of \(S_n\) formula with their \(a\) and \(d\); ok unsimplified for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2(2p+q+4q) = 40\) and \(3(2p+q+6q) = 72\) | DM1 | Uses their \(S_n\) formula from (i) |
| \(p = 5\) and \(q = 2\) [Could use \(S_n\) with \(a\) and \(d \rightarrow a=7, d=2 \rightarrow p=5, q=2\)] | A1 | Note: answers 7, 2 instead of 5, 2 gets M1A0 – must attempt to solve for M1 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $ar = 12$ and $\frac{a}{1-r} = 54$ | **B1 B1** | CAO, OE CAO, OE |
| Eliminates $a$ or $r \rightarrow 9r^2 - 9r + 2 = 0$ or $a^2 - 54a + 648 = 0$ | **M1** | Elimination leading to a 3-term quadratic in $a$ or $r$ |
| $r = \frac{2}{3}$ or $\frac{1}{3}$ hence $a \rightarrow a = 18$ or $36$ | **A1** | Needs both values |
## Question 8(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| First term $= p + q$. Difference $= q$ or last term $= p + qn$ | **B1** | Need first term and last term or common difference |
| $S_n = \frac{n}{2}(2(p+q)+(n-1)q)$ or $\frac{n}{2}(2p+q+nq)$ | **M1A1** | Use of $S_n$ formula with their $a$ and $d$; ok unsimplified for A1 |
## Question 8(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2(2p+q+4q) = 40$ and $3(2p+q+6q) = 72$ | **DM1** | Uses their $S_n$ formula from (i) |
| $p = 5$ and $q = 2$ [Could use $S_n$ with $a$ and $d \rightarrow a=7, d=2 \rightarrow p=5, q=2$] | **A1** | Note: answers 7, 2 instead of 5, 2 gets M1A0 – must attempt to solve for M1 |8
\begin{enumerate}[label=(\alph*)]
\item A geometric progression has a second term of 12 and a sum to infinity of 54 . Find the possible values of the first term of the progression.
\item The $n$th term of a progression is $p + q n$, where $p$ and $q$ are constants, and $S _ { n }$ is the sum of the first $n$ terms.
\begin{enumerate}[label=(\roman*)]
\item Find an expression, in terms of $p , q$ and $n$, for $S _ { n }$.
\item Given that $S _ { 4 } = 40$ and $S _ { 6 } = 72$, find the values of $p$ and $q$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2018 Q8 [9]}}