Interior angle of a polygon are …

Let the number of sides of polygon be The interior angles of the polygon form an A.P.

Here, a = 120^o and d = 5^o

Since Sum of interior angles of a polygon with n sides is {tex}( n - 2 ) \times 180 ^ { \circ }{/tex}

{tex}\therefore \mathrm { S } _ { n } = ( n - 2 ) \times 180 ^ { \circ }{/tex}

{tex}\Rightarrow \frac { n } { 2 } [ 2 \times 120 + ( n - 1 ) \times 5 ] = 180 n - 360{/tex}

{tex}\Rightarrow 120 n + \frac { 5 n ^ { 2 } - 5 n } { 2 } = 180 n - 360{/tex}

{tex}\Rightarrow{/tex} 240n + 5n² - 5n = 360n - 720

{tex}\Rightarrow{/tex} 5n² - 125n + 720 = 0

divide by 5, we get 

{tex}\Rightarrow{/tex} n² - 25n + 144 = 0

{tex}\Rightarrow{/tex} (n - 16) (n - 9) = 0 

{tex}\Rightarrow{/tex} n = 16 or n = 9 

But n = 16 not possible because a₁₆ = a + 15d = 120 + 15 {tex}\times{/tex} 5 = 195^o > 180^o

Therefore, number of sides of the polygon are 9.