Let the first term of the arithmetic progression be \(a = 3\) and the common difference be \(d = 2\).
The angles of the sectors form an arithmetic sequence: \(3, 5, 7, \ldots\).
The sum of the angles in a circle is 360°.
The sum of the first \(n\) terms of an arithmetic sequence is given by:
\(S_n = \frac{n}{2} (2a + (n-1)d)\)
Substitute \(a = 3\), \(d = 2\), and \(S_n = 360\):
\(\frac{n}{2} (6 + (n-1)2) = 360\)
Simplify:
\(\frac{n}{2} (6 + 2n - 2) = 360\)
\(\frac{n}{2} (2n + 4) = 360\)
\(n(n + 2) = 360\)
\(n^2 + 2n - 360 = 0\)
Solving the quadratic equation \(n^2 + 2n - 360 = 0\) using the quadratic formula:
\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
where \(a = 1\), \(b = 2\), \(c = -360\).
\(n = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-360)}}{2 \times 1}\)
\(n = \frac{-2 \pm \sqrt{4 + 1440}}{2}\)
\(n = \frac{-2 \pm \sqrt{1444}}{2}\)
\(n = \frac{-2 \pm 38}{2}\)
\(n = 18 \text{ or } n = -20\)
Since \(n\) must be positive, \(n = 18\).
