Question:
If angles of polygon are in A.P. where a=120 and d=5 then find the number of sides of polygon
Answer:
Let there are n side of the polygon.
Given that angles of the polygon are in AP and a = 120 and d = 5
Now sum of all interior angles of the polygon = (2n-4)*90
=> (n/2)*[2a + (n-1)*d] = (2n-4)*90
=> (n/2)*[2*120 + (n-1)*5] = (2n-4)*90
=> (n/2)*[240 + (n-1)*5] = (2n-4)*90
=> (n/2)*[240 + 5n- 5] = (2n-4)*90
=> n*[245 + 5n] = (2n-4)*90*2
=> 5*[49n + n2 ] = (2n-4)*180
=> 49n + n2 = (2n-4)*180/5
=> 49n + n2 = (2n-4)*36
=> 49n + n2 = 2n*36 - 4*36
=> 49n + n2 = 72n - 144
=> n2 + 49n - 72n + 144 = 0
=> n2 + 49n - 72n + 144 = 0
=> n2 - 23n + 144 = 0
=>(n - 9)*(n - 16) = 0
=> n = 9, 16
when n =16 then largest angle of the polygon = a + (n-1)*d
= 120 + (16 - 1)*5
= 120 + 15*5
= 120 +75
= 195
Which is not possible.
So number of side n =9
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