counterclockwise manner

bigmonkey31 wrote:
I'm a bit confused of how you got the "AP = n/2(2*a+(n-1)*d)". Can you explain a bit further? And how did you get "180n -360 = 5/2 (n^2+ 47n)" from that? Is it me or is 235/2 not equal to 47???? Or is it just late and I should be understanding this??!??! ahh

Sum of n terms of an arithmetic progression which has 'a' has the first term and 'd' as the common difference is

S = n/2(2*a +(n-1)*d)

In this case lowest angle is 120 degrees, and the angles are in Arithmetic progression with common difference d

a= 120 and d = 5

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

= n/2 (240 +5n -5)

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

= 5n/2 (47 +n)

and this should be equal to (n-2)*180 since (n-2)*180 is the sum of all internal angles of a convex polygon

=> 5n/2 (47 +n) = 180*(n-2)
Dividing both sides by 5/2

n(47+n) = 72*(n-2)

Rest is clear from above I suppose.

Please ask if the doubt persists!

since you know 180(n-2) = sum of interior angles
and since all the angles increase in a constant=5degrees,
you can use AP.
a+(a+d)+(a+2d)+...+(a+(n-1)d) => first try to understand this.
think about how you get average for consecutive integers. if there are five consecutive positive integers,
you can get the average by adding the first and last integer divided by 2. you can get the sum of the integers by multiplying the average by the number of integers. it's the same concept.

so then since the first angle would be a and last (a+(n-1)d) you add those up and divide by two and multiply by n.
You get Sn=n(2a+(n-1)d)/2

a, the first angle, in this case is 120. and d, the constant would be 5.
180(n-2)=n(2a+(n-1)d)/2

you have one equation with one unknown, you can solve it.