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Let the number of sides of polygon be
The interior angles of the polygon form an A.P.
Here, a = 120o and d = 5o
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 = 195o > 180o
Therefore, number of sides of the polygon are 9.

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