IIT JEE , AIEEE Forum - Mathematics , Algebra

nadeemoidu

let the no. of sides = n , sum of interior angles =

S_n = (2n-4).90

first term (angle , a) = 120^o , d = 5^o

S_n = (n/2) [2.120 + (n-1).5]

so

(2n-4).90 = (n/2)[240 + 5(n-1)]

(2n-4).180 = n(5n + 535)

solve this....u will get an =n

n² -25n + 144 = 0

(n-16)(n-9) = 0

n = 16 OR 9

a polygon cannot have a side equal to 180^o. So 16 is rejected. the answer is 9.

read this http://www.goiit.com/posts/list/algebra-angles-of-a-polygon-36194.htm