consider the sequence 1,-2,3,-4,5,-6.what is the average of the first 200 terms in the sequnce?

• ans -0.5
  we can rewrite it as (1+3+5+.....+199)-(2+4+6+....+200)
  sum of first n odd natural numbers s=1+3+5+....+(2n-1)=n^2
  sum of first n even natural numbers s=2+4+6+....+2n=n^2+n
  so acc to above sequence n^2=10000 for odd
  and n^2+n=100100 for even
  b/c both have 100 terms
  hence total sum=10000-100100=-100
  aveg=(total sum)/(total no.)
  (-100/200)=-0.5
• 10 years agoHelpfull: Yes(8) No(0)
• (-1/2)
  1-2+3-4+5-6+.......+199-200
  -1-1-1-1-1.....-1=-100
  -100/2= -1/2
• 10 years agoHelpfull: Yes(6) No(0)
• Sum of all odd numbers from 1 to 100 (50 odd numbers) =2500
  Sum of all even no from 1 to 100(50 even numbers) =2550
  As it is given first 200 terms then,
  Ans is ((2500*2)-(2550*2))/200=-0.5
• 8 years agoHelpfull: Yes(1) No(0)
• ans:0.5
  (1-2+3-4+5......199-200)/200==0.5
• 10 years agoHelpfull: Yes(0) No(0)