| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Shared terms between AP and GP |
| Difficulty | Standard +0.3 This is a straightforward system of equations problem requiring students to use basic AP and GP formulas (constant difference and constant ratio). The algebra is routine with no conceptual surprises—slightly easier than average since it's a standard 'shared terms' exercise with clear setup and mechanical solution steps. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(a + b = 2 \times \frac{3}{2}a\right) \Rightarrow b = 2a\) | B1 | SOI |
| \(18^2 = a(b+3)\) OE or 2 correct statements about \(r\) from the GP, e.g. \(r = \frac{18}{a}\) and \(b+3 = 18r\) or \(r^2 = \frac{b+3}{a}\) | B1 | SOI |
| \(324 = a(2a+3) \Rightarrow 2a^2 + 3a - 324[=0]\) or \(b^2 + 3b - 648[=0]\) or \(6r^2 - r - 12[=0]\) or \(4d^2 + 3d - 162[=0]\) | M1 | Using the correct connection between AP and GP to form a 3-term quadratic with all terms on one side |
| \((a-12)(2a+27)[=0]\) or \((b-24)(b+27)[=0]\) or \((2r-3)(3r+4)[=0]\) or \((d-6)(4d+27)[=0]\) | M1 | Solving *their* 3-term quadratic by factorisation, formula or completing the square to obtain answers for \(a\), \(b\), \(r\) or \(d\) |
| \(a = 12,\ b = 24\) | A1 | WWW. Condone extra 'solution' \(a = -13.5, b = -27\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Common difference \(d = 6\) | B1 FT | SOI. FT *their* \(\frac{a}{2}\) |
| \(S_{20} = \frac{20}{2}(2 \times 12 + 19 \times 6)\) | M1 | Using correct sum formula with *their* \(a\), *their* calculated \(d\) and 20 |
| \(1380\) | A1 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(a + b = 2 \times \frac{3}{2}a\right) \Rightarrow b = 2a$ | B1 | SOI |
| $18^2 = a(b+3)$ OE or 2 correct statements about $r$ from the GP, e.g. $r = \frac{18}{a}$ and $b+3 = 18r$ or $r^2 = \frac{b+3}{a}$ | B1 | SOI |
| $324 = a(2a+3) \Rightarrow 2a^2 + 3a - 324[=0]$ or $b^2 + 3b - 648[=0]$ or $6r^2 - r - 12[=0]$ or $4d^2 + 3d - 162[=0]$ | M1 | Using the correct connection between AP and GP to form a 3-term quadratic with all terms on one side |
| $(a-12)(2a+27)[=0]$ or $(b-24)(b+27)[=0]$ or $(2r-3)(3r+4)[=0]$ or $(d-6)(4d+27)[=0]$ | M1 | Solving *their* 3-term quadratic by factorisation, formula or completing the square to obtain answers for $a$, $b$, $r$ or $d$ |
| $a = 12,\ b = 24$ | A1 | WWW. Condone extra 'solution' $a = -13.5, b = -27$ only |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Common difference $d = 6$ | B1 FT | SOI. FT *their* $\frac{a}{2}$ |
| $S_{20} = \frac{20}{2}(2 \times 12 + 19 \times 6)$ | M1 | Using correct sum formula with *their* $a$, *their* calculated $d$ and 20 |
| $1380$ | A1 | |8 The first, second and third terms of an arithmetic progression are $a , \frac { 3 } { 2 } a$ and $b$ respectively, where $a$ and $b$ are positive constants. The first, second and third terms of a geometric progression are $a , 18$ and $b + 3$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ and $b$.
\item Find the sum of the first 20 terms of the arithmetic progression.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q8 [8]}}