A circular disc is cut into twelve sectors whose areas are in an arithmetic sequence.

The angle of the largest sector is twice the angle of the smallest sector.

Find the size of the angle of the smallest sector.

METHOD 1

If the areas are in arithmetic sequence, then so are the angles.     (M1)

\( \Rightarrow {S_n} = \frac{n}{2}(a + l) \Rightarrow \frac{{12}}{2}(\theta + 2\theta ) = 18\theta \)     M1A1

\( \Rightarrow 18\theta = 2\pi \)     (A1)

\(\theta = \frac{\pi }{9}\)     (accept \(20^\circ \))     A1

[5 marks] 

METHOD 2

\({{\text{a}}_{12}} = 2{a_1}\)     (M1)

\(\frac{{12}}{2}({a_1} + 2{a_1}) = \pi {r^2}\)     M1A1

\(3{a_1} = \frac{{\pi {r^2}}}{6}\)

\(\frac{3}{2}{r^2}\theta = \frac{{\pi {r^2}}}{6}\)     (A1)

\(\theta = \frac{{2\pi }}{{18}} = \frac{\pi }{9}\)     (accept \(20^\circ \))     A1

[5 marks] 

METHOD 3

Let smallest angle = a , common difference = d

\(a + 11d = 2a\)     (M1)

\(a = 11d\)     A1

\({S_n} = \frac{{12}}{2}(2a + 11d) = 2\pi \)     M1

\(6(2a + a) = 2\pi \)     (A1)

\(18a = 2\pi \)

\(a = \frac{\pi }{9}\)     (accept \(20^\circ \))     A1

[5 marks]

Stronger candidates had little problem with this question, but a significant minority of weaker candidates were unable to access the question or worked with area and very quickly became confused. Candidates who realised that the area of each sector was proportional to the angle usually gained the correct answer.