| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Form and solve quadratic in parameter |
| Difficulty | Moderate -0.3 This is a straightforward geometric progression problem requiring students to use the property that consecutive terms have a constant ratio, leading to a standard quadratic equation. The sum to infinity calculation in part (b) is routine application of a formula. While it requires careful algebraic manipulation, it's slightly easier than average as it follows a well-practiced procedure with no novel insight needed. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{5p}{2p+6} = \frac{8p+2}{5p}\) | M1 | OE. Setting up a valid relationship in terms of \(p\) |
| \(9p^2 - 52p - 12 = 0\) | DM1 | OE. Simplifying to a 3 term quadratic equation, only condone sign errors |
| \([(9p+2)(p-6) = 0]\) leading to \(p = -\frac{2}{9}\) and \(6\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 2\left(-\frac{2}{9}\right) + 6\left[= \frac{50}{9}\right]\) | \*M1 | FT *their* \(-\frac{2}{9}\), allow any negative non-integer |
| \(r = -\frac{10}{9} \div \frac{50}{9}\left[= -\frac{1}{5}\right]\) | \*M1 | FT *their* \(-\frac{2}{9}\), allow any negative non-integer |
| \(S_\infty = \frac{50}{9} \div \left(1--\frac{1}{5}\right) = \frac{125}{27}\) | DM1 A1 | Can only get DM1 if \( |
| 4 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{5p}{2p+6} = \frac{8p+2}{5p}$ | M1 | OE. Setting up a valid relationship in terms of $p$ |
| $9p^2 - 52p - 12 = 0$ | DM1 | OE. Simplifying to a 3 term quadratic equation, only condone sign errors |
| $[(9p+2)(p-6) = 0]$ leading to $p = -\frac{2}{9}$ and $6$ | A1 | |
| | **3** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 2\left(-\frac{2}{9}\right) + 6\left[= \frac{50}{9}\right]$ | \*M1 | FT *their* $-\frac{2}{9}$, allow any negative non-integer |
| $r = -\frac{10}{9} \div \frac{50}{9}\left[= -\frac{1}{5}\right]$ | \*M1 | FT *their* $-\frac{2}{9}$, allow any negative non-integer |
| $S_\infty = \frac{50}{9} \div \left(1--\frac{1}{5}\right) = \frac{125}{27}$ | DM1 A1 | Can only get DM1 if $|r| < 1$. Accept AWRT 4.63 |
| | **4** | |5 The first, second and third terms of a geometric progression are $2 p + 6,5 p$ and $8 p + 2$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the possible values of the constant $p$.
\item One of the values of $p$ found in (a) is a negative fraction.
Use this value of $p$ to find the sum to infinity of this progression.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q5 [7]}}