SOLUTION: The interior angles of a polygon are in A.P. if the smallest angle is 120� and common difference is 5� then the number of sides in polygon is ?

Question 983093: The interior angles of a polygon are in A.P. if the smallest angle is 120� and common difference is 5� then the number of sides in polygon is ?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
first term: 120
second term: 120+5
third term: 120+5*2
...
...
...
nth term: 120+5*(n-1)
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Sum of the arithmetic terms

Sn = n*(a1 + an)/2

Sn = n*(120 + 120+5*(n-1))/2

Sn = n*(120 + 120+5n-5)/2

Sn = n*(5n+235)/2

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Sum of angles in a polygon

S = 180(n-2)

n*(5n+235)/2 = 180(n-2) ... replace S with the arithmetic sum found above

n*(5n+235) = 2*180(n-2)

5n^2+235n = 360n-720

5n^2+235n - 360n+720 = 0

5n^2-125n+720 = 0

Now use the quadratic formula to solve for n. I'll let you do this part.

If you get any n values that are negative or not whole numbers, then ignore them. The value of n is going to be some positive whole number.
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