the interior angles of an octagon ABCDEFGH are in AP. If the largest and the second largest have an average of 153 find the average of the least two.

• 117
  
  sum of interior angles = 4*360-360=1080 deg
  if a is smallest angle and d is difference between two angles in an AP, then
  average of largest and second largest angle = (a+6da+7d)/2= 153
  2a+13d= 306
  sum of angles = (8/2)*(2a+7d)= 1080 or 2a+7d= 270
  
  solving, a= 114, d=6
  average of least two angles= (114+120)/2= 117
• 11 years agoHelpfull: Yes(24) No(2)
• 117...sum of interior angles=(2n-4)right angles...this is one condn.....other condn is given in question...2 eq 2 var..
• 11 years agoHelpfull: Yes(2) No(1)
• 117
  x+(x+d)+(x+2d)+(x+3d)+(x+4d)+(x+5d)+(x+6d)+(x+7d)=1080
  8x+28d=1080
  2x+7d=270...................................................................................................eq1
  and he largest and the second largest have an average of 153
  
  [(x+6d)+(x+7d)]/2=153
  (x+6d)+(x+7d)=306
  2x+13d=306.................................eq2
  solve above 2 eq
  d=7
  average of the least two.=117
• 9 years agoHelpfull: Yes(0) No(0)