Consider the sequence 1,-2, 3,-4, 5,-6, ….. , what is the average of the first 2000 terms of the sequence?

• Ans is -1/2
  =-(1+1+1.....upto 1000 trms)/2000
  =-1/2
• 10 years agoHelpfull: Yes(21) No(1)
• (1,3,5,..1999) here n=1000
  s1=n/2(2a+(n-1)d)=10000000
  (-2,-4,-6...,-2000) here n=1000
  s2=-1001000
  so avg term= s1+s2/2000term
  =-1000/2000
  =-1/2 ans.
  
• 10 years agoHelpfull: Yes(4) No(0)
• ans.=-1/2
  -1-1-1....(1000th)/2000
• 10 years agoHelpfull: Yes(2) No(0)
• 1-2=-1
  3-4=-1
  .
  .
  .
  .
  -1,1000 times
  so ans=-1000/2000
  =-1/2
• 10 years agoHelpfull: Yes(2) No(0)
• @KARTHIKEYAN
  TOTAL 2000 TERMS
  JUST TAKE IT THE PAIR SEQUENCE EX:(1,-2)
  SO 1000 PAIRS IN 2000 TERMS
  
• 10 years agoHelpfull: Yes(2) No(1)
• ans is -1/2..
  
• 10 years agoHelpfull: Yes(2) No(0)
• Ans is -1/2
  =-(1+1+1.....upto 100 trms)/200
  =-1/2
• 10 years agoHelpfull: Yes(1) No(0)
• 1+3+5...1000 terms + -2-4-6..1000 terms
  1000/2[(2+(999*2))+2*(999*-2)]
  =-998000
• 10 years agoHelpfull: Yes(0) No(6)
• pls explain in the ans karthi
  
• 10 years agoHelpfull: Yes(0) No(0)
• ( on comininig 2 term eg((1,-2)(3,-4)...) we get -1 for 2000 term sum would be
  -1000) av =(-1000)/2000
  =-1/2
• 10 years agoHelpfull: Yes(0) No(0)
• first sequence is 1,3,5...... (1000 terms )
  second sequence is -2,-4.-6........ (1000 terms)
  use formula to find the sum of series =(n/2)(2a+(n-1)d)
  answere = -1/2
• 10 years agoHelpfull: Yes(0) No(0)
• @Karthikeyank:
  (-1/2) is incorrect.
  Average of a series is always positive.
  ans is (+1/2)
• 10 years agoHelpfull: Yes(0) No(1)
• sum=500*(2+4)+500*(-4-4)/2000=-.5
• 10 years agoHelpfull: Yes(0) No(0)
• ans is -0.5
• 9 years agoHelpfull: Yes(0) No(0)
• see guyzz, here 1,-2,3,-4,.....,200
  now if we know that what is the answer of first 10 terms then we can calculate the for 200 terms.
  here summing first 10 terms
  1-2+3-4+5-6+7-8+9-10
  =-1-1-1-1-1
  =-5
  here for 10 terms answer is half 10/2=-5
  so that for 200=200/2=-100
  and for average total by total numbers=-100/200=-.5
• 9 years agoHelpfull: Yes(0) No(0)