The interior angles of a polygon with n sides, are in an A.P. with common difference 6°. If the largest interior angle of the polygon is 219°, then n is equal to _______.

Solution

Maths sir · 4 steps

Write the angle sum two ways, once as an A.P. sum and once as (n − 2) × 180°, and set them equal.

1. Set up the A.P. The largest angle, 219°, is the last term. Counting back n-1 steps of 6°, the smallest angle is 219 - 6(n-1) = 225 - 6n degrees.
2. Sum of the A.P. The sum is n/2[(225 - 6n) + 219] = n(222 - 3n) degrees. This must equal the polygon's interior angle sum, (n-2) × 180°.

   n(222 - 3n) = 180(n-2) ⇒ 3n² - 42n - 360 = 0 ⇒ (n-20)(n+6) = 0

3. Check it fits. With n = 20 the angles run from 105° to 219°, and their sum is 10 × 324 = 3240° = 18 × 180°. An angle above 180° is fine, because the polygon is concave.
4. So. n = -6 is rejected, so n = 20.

The common slip: 14. Using 180n for the angle sum instead of (n-2) × 180° gives 14, not 20.

The idea to keep. For polygon angles in an A.P., equate the A.P. sum to (n-2) × 180°. Then check that the smallest angle is positive before you accept the value of n.

Answer from the NTA final answer key. Solution by PrepWiser.