What is the sum of 1-2+3-4+5.......-98+99 ?

• 1-2+3-4+5.......-98+99
  =(1-2)+(3-4)+(5-6)+....+(97-98)+99
  =(-1-1-1....49 times)+99
  =-49+99
  =50
• 10 years agoHelpfull: Yes(59) No(0)
• 1-2+3-4...............-98+99=1+(-2+3)+..........+(-98+99)
  1+(1+1+.......49times)=1+49=50
  ans. 50
• 10 years agoHelpfull: Yes(4) No(0)
• two separate APs are formed
  ((1+3+.....+99)-(2+4+.......+98))=50
• 10 years agoHelpfull: Yes(2) No(0)
• ans is 50
  (1+3+.....99)-(2+4+....98)
  now use sum of airthmetic progression
  
• 10 years agoHelpfull: Yes(2) No(1)
• For every group of two numbers from the start the result is (-1). Like +1-2 is -1; +3-4 is -1; +5-6 is -1; and so on .
  Up to 98, there are 49 groups of pairs of numbers, all of them together result - 49.
  So, teh final result is -49+ the last term 99 = 50
  
  50 is the answer
• 10 years agoHelpfull: Yes(1) No(1)
• We can rearrange it as
  1+(-2+3)+(-4+5)+...........(-98+99)=1+(1*49)=50Ans.
• 10 years agoHelpfull: Yes(1) No(0)
• AP(1)=(1+3+5+....+99)
  AP(2)=(2+4.....+98).
  ANS=SUM OF AP(1)+SUM OF AP(2)
• 10 years agoHelpfull: Yes(1) No(1)
• ans is 50
  1-2=-1
  3-4=-1
  series repeat for 48 times
  last 99-49=50
• 10 years agoHelpfull: Yes(0) No(0)
• 50
  
  1-2=(-1)
  3-4=(-1)
  .
  .
  .
  97-98=(-1)
  there r 24 pairs(till 98)which gives the sum = -24
  and 99-24=50
• 10 years agoHelpfull: Yes(0) No(1)
• ooops!!!
  mistakenly i wrote 24 terms....there are 49 terms actually..
• 10 years agoHelpfull: Yes(0) No(1)
• (1+2+3+4....+99)-2(2+4+6+8...+98)
  =(99*100/2)-4(1+2+3+4...+49)
  =4950-4900
  =50
  ans:50
• 9 years agoHelpfull: Yes(0) No(0)
• (1+3+5+......)-(2+4+6+....)
  in odd series number of terms=50
  in even series number of terms =49
  sum of odd series=n^2
  sum of even series=n(n+1)
  ans=(50)^2-(49*50)=50
  
• 9 years agoHelpfull: Yes(0) No(0)