| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | GP with trigonometric terms |
| Difficulty | Standard +0.8 Part (a)(i) is trivial recall of the sum to infinity formula. Part (a)(ii) requires recognizing that sin 30° = 1/2 and applying the same formula. Part (b) is more challenging: students must set up S_∞ = 1/(1-cos θ) = 2-√2, rearrange to find cos θ, then solve the trigonometric equation for the smallest positive radian value. The combination of geometric series with trigonometry and the requirement for exact radian answers elevates this above a standard textbook exercise, though it remains accessible with systematic working. |
| Spec | 1.04j Sum to infinity: convergent geometric series |r|<11.05g Exact trigonometric values: for standard angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(\frac{a}{1-r}\) | M1 | |
| \(S_\infty = \frac{1}{1-\frac{1}{2}} = 2\) | A1 | Obtains \(2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Deduces \(a=\frac{1}{2}\) and \(r=\frac{1}{2}\), or deduces \(\sum_{n=1}^{\infty}(\sin 30°)^n =\) their part (a)(i) \(-1\), or deduces the answer is half of their answer in part (a)(i) | M1 | \(\sum_{n=1}^{\infty}(\sin 30°)^n = \frac{1}{2}+\frac{1}{4}+\ldots = \frac{\frac{1}{2}}{1-\frac{1}{2}}\) |
| Obtains \(1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Forms equation \(\frac{a}{1-r} = 2-\sqrt{2}\); condone \(\frac{\cos\theta}{1-\cos\theta}=2-\sqrt{2}\) or numerical value for \(a\) where \(a>0\) | M1 | \(\sum_{n=0}^{\infty}(\cos\theta)^n = 2-\sqrt{2}\) |
| Uses \(a=1\) and \(r=\cos\theta\) | B1 | |
| Obtains \(r = 1-\frac{1}{2-\sqrt{2}}\) or \(-\frac{\sqrt{2}}{2}\), or \(\cos\theta = 1-\frac{1}{2-\sqrt{2}}\) or \(-\frac{\sqrt{2}}{2}\) (ACF) | A1 | \(1-\cos\theta = \frac{1}{2-\sqrt{2}} \Rightarrow \cos\theta = 1-\frac{1}{2-\sqrt{2}}\) |
| Deduces \(\theta = \frac{3\pi}{4}\) | R1 |
## Question 12(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $\frac{a}{1-r}$ | M1 | |
| $S_\infty = \frac{1}{1-\frac{1}{2}} = 2$ | A1 | Obtains $2$ |
---
## Question 12(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Deduces $a=\frac{1}{2}$ and $r=\frac{1}{2}$, or deduces $\sum_{n=1}^{\infty}(\sin 30°)^n =$ their part (a)(i) $-1$, or deduces the answer is half of their answer in part (a)(i) | M1 | $\sum_{n=1}^{\infty}(\sin 30°)^n = \frac{1}{2}+\frac{1}{4}+\ldots = \frac{\frac{1}{2}}{1-\frac{1}{2}}$ |
| Obtains $1$ | A1 | |
---
## Question 12(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Forms equation $\frac{a}{1-r} = 2-\sqrt{2}$; condone $\frac{\cos\theta}{1-\cos\theta}=2-\sqrt{2}$ or numerical value for $a$ where $a>0$ | M1 | $\sum_{n=0}^{\infty}(\cos\theta)^n = 2-\sqrt{2}$ |
| Uses $a=1$ and $r=\cos\theta$ | B1 | |
| Obtains $r = 1-\frac{1}{2-\sqrt{2}}$ or $-\frac{\sqrt{2}}{2}$, or $\cos\theta = 1-\frac{1}{2-\sqrt{2}}$ or $-\frac{\sqrt{2}}{2}$ (ACF) | A1 | $1-\cos\theta = \frac{1}{2-\sqrt{2}} \Rightarrow \cos\theta = 1-\frac{1}{2-\sqrt{2}}$ |
| Deduces $\theta = \frac{3\pi}{4}$ | R1 | |12
\begin{enumerate}[label=(\alph*)]
\item A geometric sequence has first term 1 and common ratio $\frac { 1 } { 2 }$\\
12 (a) (i) Find the sum to infinity of the sequence.\\
12 (a) (ii) Hence, or otherwise, evaluate
$$\sum _ { n = 1 } ^ { \infty } \left( \sin 30 ^ { \circ } \right) ^ { n }$$
12
\item Find the smallest positive exact value of $\theta$, in radians, which satisfies the equation
$$\sum _ { n = 0 } ^ { \infty } ( \cos \theta ) ^ { n } = 2 - \sqrt { 2 }$$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2022 Q12 [8]}}