Question

The interior angles of a polygon are in A.P. If the smallest angle be $120 ^ { \circ }$ and the common difference be 5^o, then the number of sides is.

• A. 8
• B. 10
• C. 9
• D. 6

Answer 1

Answer: (C)

9

Answer 2

Let the number of sides of the polygon be $n$ .

Then the sum of interior angles of the polygon

$= ( 2 n - 4 ) \frac { \pi } { 2 } = ( n - 2 ) \pi$

Since the angles are in A.P. and $a = 120 ^ { \circ } , d = 5$,

therefore $\frac { n } { 2 } [ 2 \times 120 + ( n - 1 ) 5 ] = ( n - 2 ) 180$

⇒ $n ^ { 2 } - 25 n + 144 = 0$ ⇒$( n - 9 ) ( n - 16 ) = 0$⇒ $n = 9,16$

But $n = 16$ gives $T _ { 16 } = a + 15 d = 120 ^ { \circ } + 15.5 ^ { \circ } = 195 ^ { \circ }$, which is impossible as interior angle cannot be greater than $180 ^ { \circ }$ . Hence $n = 9$ .