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 \( n = 20 \).

The net current flowing in the given circuit is ___ A.

If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .