60,48,38,28,24,20,18.Find the number in the series which is defective.
9.1,-2,3,-4,5,-6.......Find the average upto 200 terms.

• 18+2=20;20+4=24;24+6=30;30+8=38;38+10=48;48+12=60
  so defective is 28 it must be 30.
  For second series avg=(-1-1...upto 100 term)/200=-1/2
• 11 years agoHelpfull: Yes(23) No(0)
• 18+2=20;
  20+4=24;
  24+6=30;
  30+8=38;
  38+10=48;
  48+12=60
  so defective is 28,which would be 30....
  for 2nd...
  1-2+3-4+5-6+7-8........upto 200 terms i.e.(-200).
  so sum would be...-100/200=-1/2
• 11 years agoHelpfull: Yes(8) No(0)
• For First
  18+2=20;
  20+4=24;
  24+6=30;
  30+8=38;
  38+10=48;
  48+12=60
  Defective is 28, It should be 30
  
  for 2nd ...i.e. 1,-2,3,-4,5,-6,...........-200.
  For Average ((1+3+5+7.....+199)-(2+4+6+8+.......+200))/200
  =((10000)-(10100))/200
  =(-1/2)
  =(-100)/200
  
• 11 years agoHelpfull: Yes(2) No(0)
• in first series difference is in A.P.
  so 18+2 =20
  20+4= 24
  24+6 =30 so 2 is the defective.
  for second avg will be = -0.5
  30+8= 38
  38+10=48
• 11 years agoHelpfull: Yes(1) No(0)
• difference of the series is:12,10,8,6,4.... so 28 is DEFECTIVE!!!
• 11 years agoHelpfull: Yes(0) No(0)
• Sum of 'n' first odd numbers is n^2
  Sum of 'n' first even numbers is n(n+1)
  =(100^2-(100*101))/200
  =-1/2
• 11 years agoHelpfull: Yes(0) No(0)
• 2nd question: (1+3+5+7+.....199)-(2+4+6+8+.....200)/2
  for sum of (1+3+5+...)= 2*n-1=(2*200)-1=399
  for sum of (2+4+6+...)=2*n=2*200=400
  so 399-400/2=-1/2
• 11 years agoHelpfull: Yes(0) No(0)