2022 H2 Maths Paper 1 Q9 — Worked Solution

Part (a)

3 marks

The arithmetic series has first term $a$ and common difference $d$ . Its 1st, 3rd and 15th terms are

T_{1} = a,\quad T_{3} = a + 2d,\quad T_{15} = a + 14d

Since these are the 1st, 2nd and 3rd terms of a geometric series, the square of the middle term equals the product of the outer two:

T_{3}^{2} = T_{1}T_{15}

\Rightarrow (a + 2d)^{2} = a(a + 14d)

\Rightarrow a^{2} + 4ad + 4d^{2} = a^{2} + 14ad

\Rightarrow 4d^{2} - 10ad = 0

\Rightarrow 2d(2d - 5a) = 0

Since $d \neq 0$ ,

2d - 5a = 0

To check your answer: with

a = 2

,

d = 5

:

T_{1} = 2

,

T_{3} = 12

,

T_{15} = 72

; ratios

\frac{12}{2} = 6

and

\frac{72}{12} = 6

agree ✓

$\therefore$ $d = \frac{5a}{2}$ .

Part (b)(i)

3 marks

Approach

with a two-sided sum-to-infinity expression, the half-angle identity turns $\frac{\sin\theta}{1 + \cos\theta}$ directly into a single tangent.

The geometric series has first term $\sin\theta$ and common ratio $- \cos\theta$ . Since $0 < \theta < \frac{\pi}{2}$ , $0 < \cos\theta < 1$ , so $\left| - \cos\theta \right| < 1$ and the sum to infinity exists:

S_{\infty} = \frac{\sin\theta}{1 - \left( - \cos\theta \right)} = \frac{\sin\theta}{1 + \cos\theta}

Using the half-angle identity $\tan\left( \frac{\theta}{2} \right) = \frac{\sin\theta}{1 + \cos\theta}$ ,

S_{\infty} = \tan\left( \frac{\theta}{2} \right)\ \ \left( \text{shown} \right)

$\therefore$ $k = \frac{1}{2}$ .

Part (b)(ii)

2 marks

S_{7} = \frac{43\sqrt{3}}{128}

Explanation: at $\theta = \frac{\pi}{3}$ , the first term is $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and the common ratio is $- \cos\frac{\pi}{3} = - \frac{1}{2}$ . Using $S_{n} = \frac{a\left( 1 - r^{n} \right)}{1 - r}$ with $n = 7$ ,

S_{7} = \frac{\frac{\sqrt{3}}{2}\left( 1 - \left( - \frac{1}{2} \right)^{7} \right)}{1 - \left( - \frac{1}{2} \right)}

= \frac{\frac{\sqrt{3}}{2} \cdot \frac{129}{128}}{\frac{3}{2}}

= \frac{\sqrt{3}}{2} \times \frac{129}{128} \times \frac{2}{3}

= \frac{129\sqrt{3}}{384}

Since $129 = 3 \times 43$ and $384 = 3 \times 128$ , this simplifies to

S_{7} = \frac{43\sqrt{3}}{128}

To check your answer:

\frac{43\sqrt{3}}{128} \approx 0.582

, matching a direct numerical sum of the first 7 terms ✓

2022 H2 Maths Paper 1, Question 9

Part (a)

Part (b)(i)

Part (b)(ii)

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