| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Relationship between two GPs |
| Difficulty | Standard +0.8 This question requires understanding of sum to infinity formula for GPs, algebraic manipulation across two simultaneous conditions, and solving a system involving both the sum relationship and term equality. While the individual components are standard A-level content, combining these constraints to eliminate variables and reach the final expression requires more sophisticated algebraic reasoning than typical textbook exercises. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = \frac{a}{1-r}\), \(2S = \frac{a}{1-R}\) | B1 | SOI at least one correct |
| \(\frac{2a}{1-r} = \frac{a}{1-R}\) | M1 | SOI |
| \(2 - 2R = 1 - r \rightarrow r = 2R - 1\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(ar^2 = aR \rightarrow (a)(2R-1)^2 = R(a)\) | *M1 | |
| \(4R^2 - 5R + 1\ (=0) \rightarrow (4R-1)(R-1)\ (=0)\) | DM1 | Allow use of formula or completing the square |
| \(R = \frac{1}{4}\) | A1 | Allow \(R = 1\) in addition |
| \(S = \frac{2a}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(ar^2 = aR \rightarrow (a)r^2 = \frac{1}{2}(r+1)(a)\) | *M1 | Eliminating 1 variable |
| \(2r^2 - r - 1\ (=0) \rightarrow (2r+1)(r-1)\ (=0)\) | DM1 | Allow use of formula or completing the square. Must solve a quadratic |
| \(r = -\frac{1}{2}\) | A1 | Allow \(r = 1\) in addition |
| \(S = \frac{2a}{3}\) | A1 |
## Question 8(a):
| $S = \frac{a}{1-r}$, $2S = \frac{a}{1-R}$ | B1 | SOI at least one correct |
|---|---|---|
| $\frac{2a}{1-r} = \frac{a}{1-R}$ | M1 | SOI |
| $2 - 2R = 1 - r \rightarrow r = 2R - 1$ | A1 | AG |
---
## Question 8(b):
| $ar^2 = aR \rightarrow (a)(2R-1)^2 = R(a)$ | *M1 | |
|---|---|---|
| $4R^2 - 5R + 1\ (=0) \rightarrow (4R-1)(R-1)\ (=0)$ | DM1 | Allow use of formula or completing the square |
| $R = \frac{1}{4}$ | A1 | Allow $R = 1$ in addition |
| $S = \frac{2a}{3}$ | A1 | |
**Alternative method:**
| $ar^2 = aR \rightarrow (a)r^2 = \frac{1}{2}(r+1)(a)$ | *M1 | Eliminating 1 variable |
|---|---|---|
| $2r^2 - r - 1\ (=0) \rightarrow (2r+1)(r-1)\ (=0)$ | DM1 | Allow use of formula or completing the square. Must solve a quadratic |
| $r = -\frac{1}{2}$ | A1 | Allow $r = 1$ in addition |
| $S = \frac{2a}{3}$ | A1 | |
---8 A geometric progression has first term $a$, common ratio $r$ and sum to infinity $S$. A second geometric progression has first term $a$, common ratio $R$ and sum to infinity $2 S$.
\begin{enumerate}[label=(\alph*)]
\item Show that $r = 2 R - 1$.\\
It is now given that the 3rd term of the first progression is equal to the 2nd term of the second progression.
\item Express $S$ in terms of $a$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q8 [7]}}