## Question 15:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses common ratio $\dfrac{5+2\sin\theta}{12\cos\theta} = \dfrac{6\tan\theta}{5+2\sin\theta}$ | M1 | Key step using ratio $\frac{a_2}{a_1} = \frac{a_3}{a_2}$ |
| Cross multiplies and uses $\tan\theta \times \cos\theta = \sin\theta$: $(5+2\sin\theta)^2 = 6 \times 12\sin\theta$ | dM1 | Cross multiplies and uses $\tan\theta\cos\theta = \sin\theta$ |
| $25 + 20\sin\theta + 4\sin^2\theta = 72\sin\theta \Rightarrow 4\sin^2\theta - 52\sin\theta + 25 = 0$ | A1* | Proceeds to given answer including "$= 0$" with no errors and sufficient working |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\sin^2\theta - 52\sin\theta + 25 = 0 \Rightarrow \sin\theta = \frac{1}{2}\left(, \frac{25}{2}\right)$ | M1 | |
| $\theta = \dfrac{5\pi}{6}$ | A1 | |

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts a value for either $a$ or $r$: e.g. $a = 12\cos\theta = 12\times -\dfrac{\sqrt{3}}{2}$ or $r = \dfrac{5+2\sin\theta}{12\cos\theta} = \dfrac{5+2\times\frac{1}{2}}{12\times-\frac{\sqrt{3}}{2}}$ | M1 | |
| $"a" = -6\sqrt{3}$ and $"r" = -\dfrac{1}{\sqrt{3}}$ | A1 | |
| Uses $S_\infty = \dfrac{a}{1-r} = \dfrac{-6\sqrt{3}}{1+\dfrac{1}{\sqrt{3}}}$ | dM1 | |
| Rationalises denominator: $S_\infty = \dfrac{-6\sqrt{3}}{\dfrac{1+\frac{1}{\sqrt{3}}}{}} = \dfrac{-18}{\sqrt{3}+1}\times\dfrac{\sqrt{3}-1}{\sqrt{3}-1}$ | ddM1 | |
| $S_\infty = 9(1-\sqrt{3})$ | A1 | |

# Question 15 (GP/Trigonometry):

## Part (a) - Alternative Method:

| Working | Mark | Guidance |
|---------|------|----------|
| $u_2 = \frac{u_1 \times u_3}{u_2} \Rightarrow 5 + 2\sin\theta = \frac{12\cos\theta \times 6\tan\theta}{5 + 2\sin\theta}$ | M1 | Expresses 2nd and 3rd terms in terms of first term and common ratio, eliminates $r$ |
| $(5 + 2\sin\theta)^2 = 72\sin\theta$ | dM1 | Multiplies up and uses $\tan\theta \times \cos\theta = \sin\theta$ |
| $25 + 20\sin\theta + 4\sin^2\theta = 72\sin\theta$ | | |
| $\Rightarrow 4\sin^2\theta - 52\sin\theta + 25 = 0$ * | A1 | Proceeds to given answer including "$= 0$" with no errors and sufficient working |

## Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| Solve $4\sin^2\theta - 52\sin\theta + 25 = 0$ | M1 | Must be clear they have found $\sin\theta$ and not just $x$ from $4x^2 - 52x + 25 = 0$ |
| $\theta = \dfrac{5\pi}{6}$ and no other values | A1 | Unless $\dfrac{5\pi}{6}$ clearly selected here and not in (c); minimum requirement: $\sin\theta = \frac{1}{2}$, $\theta = \frac{5\pi}{6}$; do **not** allow $150°$ for $\frac{5\pi}{6}$ |

## Part (c):

| Working | Mark | Guidance |
|---------|------|----------|
| Attempt value for $a$ or $r$ using their $\theta$; e.g. $a = 12\cos\theta = \left(12 \times -\frac{\sqrt{3}}{2}\right)$ or $r = \frac{5 + 2\sin\theta}{12\cos\theta}$ | M1 | For attempting a value (exact or decimal) for either $a$ or $r$ using their $\theta$ |
| $a = -6\sqrt{3}$ and $r = -\dfrac{1}{\sqrt{3}}$ | A1 | May be left unsimplified; requires $\sin\theta = \frac{1}{2}$, $\cos\theta = -\frac{\sqrt{3}}{2}$, $\tan\theta = -\frac{\sqrt{3}}{3}$ |
| $S_\infty = \dfrac{a}{1-r} = \dfrac{-6\sqrt{3}}{1 + \dfrac{1}{\sqrt{3}}}$ | dM1 | Uses both values of $a$ and $r$ with $S_\infty = \frac{a}{1-r}$; requires $|r| < 1$; depends on first M mark |
| Rationalises denominator; denominator of form $p \pm q\sqrt{3}$ | ddM1 | e.g. $\dfrac{k}{p + q\sqrt{3}} \times \dfrac{p - q\sqrt{3}}{p - q\sqrt{3}}$; depends on both previous M marks |
| $S_\infty = 9(1 - \sqrt{3})$ | A1 | Full marks available for use of $\theta = 150°$ in (c) |

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