We are given that the interior angles are in arithmetic progression (A.P.)
with a common difference of 6° and the largest angle is 219°. The sum of the interior angles of an \( n \)-sided polygon is given by: \[ \frac{n}{2} \left( 2a + (n-1) \times 6 \right) = (n-2) \times 180 \] where \( a \) is the first angle. Simplifying: \[ an + 3n^2 - 3n = (n-2) \times 180 \] Now, using the condition that the largest interior angle is 219°, we have: \[ a + (n-1) \times 6 = 219 \] which simplifies to: \[ a = 225 - 6n \] Substitute this value of \( a \) into the sum equation: \[ (225 - 6n) + 3n^2 - 3n = (n-2) \times 180 \] Solving the resulting quadratic equation gives \( n = 20 \).
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: