satty lavanya

hiiiiii friends.... i have 1 sum... as i got the answer 3 bt the actual answeris 8 .

as the question follows.....

q: the angles of a polygon are in arithmetic progression .if the smallest angle is 100 degrees and the common difference is 10 degrees,then how many sides does the polygon have?

as i had done in the following way....

in A.P sum on n terms is n/2[2a+(n-1)d] as the first term is the least term a=100 d=10

as in polygon .....sum of the angles =360 as i substituted i got n=3....

is it not the correct way?

According to the question you have provided, it seems that you have considered only the internal angles of the polygon.

Any external angle e= 180-i where i is the corresponding internal angle.

If the internal angles of polygon are in AP then external too would be in AP.(I would leave this to you to verify)

Now when I consider the external angles of the polygon, then

the maximum external angle e would be the corresponding minimum internal angle i,

in our case, it would be against 100 degs.

hence a=180-100=80degs.

Since we are approaching from maximum term, the common difference now would be -10 degs.

Sum of all the internal angles of a polygon = Sum of all the external angles of a polygon, on a 2d rectangular surface=360 degs.

(n/2)(2a+(n-1)d)=360

n(2*(80)+(n-1)*(-10))=720

160n+(n^2-n)*(-10)=720

160n+(-10n^2+10n)=720

-10n^2+170n=720

10n^2-170n+720=0

After Calculating the roots,

n=9 or 8.

When n=9,

Internal angles would vary from minimum internal angle 100 degs to 100+(9-1)*10

=100+80 =180.

This is not possible in a polygon which is defined above.

When n=8, the Internal angles would vary from 100 to 100+(8-1)*10

=100+70=170.

This is acceptable, hence n=8.