1-2+3-4+5-6+...................-98+99=??

• ANS IS 50
  
  taking + digits and - digits separately
  for (1,3,5,7......97,99) and by applying AP WE GET SUM AS 2500
  FOR (2,4,6....,96,98) AND BY APPLYING AP WE GET -2450
  SO TAKING 2500-2450=50 ANS
• 11 years agoHelpfull: Yes(31) No(0)
• 1-2+3-4+5-6+........-98+99
  = (1-2)+(3-4)+(5-6)+.....+(97-98)+99
  = -1-1-1-1.............-1(49th)+99
  = -49+99 (as every bracket gives -1 and total no. of bracket is 49 so -49)
  = 50 ans.
• 11 years agoHelpfull: Yes(15) No(0)
• 1-2+3-4+5-6+ . . -98+99
  = (1+3+5+7...+99) - (2+4+6+8....+98)
  
  According to AP, if we know the first and last term and the total numbers then,
  AP = n/2(a+a(n)) where a is the first number and a(n) is the last number.
  
  Total odd nos. between 1 to 99 = 50
  Total even nos. between 2 to 98 = 49
  
  => (1+3+5+7...+99) - (2+4+6+8....+98) = {50/2(1+99) - 49/2(2+98)}
  => 25*100 - 50*49
  => 2500 - 2450
  = 50
  
  Answer is 50.
• 11 years agoHelpfull: Yes(13) No(0)
• We have to divide this sum into two parts :
  
  
  1+3+5+7......+99
  
  And
  
  +(-2)+(-4)+(-6)+(-8).......+(-98)
  
  Using the A.P formula, we get,
  
  L= a+(n-1)d
  
  where L= Last term , a= First term , n= Number of terms , d= Difference
  
  In first case,1+3+5+7......+99
  
  a=1, L=99, d=2
  
  So, 99=1+(n-1)2
  => (n-1)2 =98
  => n-1=49
  => n=50
  
  So, the sum of 1+3+5+7......+99= n/2(a+L)=50/2(1+99)=25×100=2500
  In the 2nd case,+(-2)+(-4)+(-6)+(-8).......+(-98)= -(2+4+6+8.....+98)
  
  Here, a=2, d=2 and L=98
  
  So, L=a+(n-1)d
  => 98=2+(n-1)2
  => (n-1)2=96
  => n-1=48
  => n=49
  
  So the sum of (2+4+6+8.....+98)=n/2(a+L)=49/2(2+98)=(49×100)/2=49×50=2450
  
  Therefore, 1-2+3-4+5-6+...................-98+99=(1+3+5+7......+99)-((2+4+6+8.....+98)=2500-2450=50
  
• 11 years agoHelpfull: Yes(11) No(0)
• (1+3+5+.....+99)-(2+4+6+........+98)
  =50/2 (1+99)-49/2 (2+98)
  =2500-2450
  =50.........ri8...........
• 11 years agoHelpfull: Yes(3) No(0)
• taking 1 appart
  (-2+3)+(-4+5)+..... ==> (1)+(1)+.....
  1 to 99 means 99 digits..
  neglecting 1 for a while
  there are 98 digits...
  98/2=49
  49*1+1= 50
  so, 50 is the ans
• 11 years agoHelpfull: Yes(2) No(0)
• 1st series::
  1,3,5,........99
  In A.P with d=2 and a=1
  To find Sn find n?
  As we know,An= a+(n-1)d
  So,here 99=1+(n-1)2
  2n-1=99 or 2n=100
  n=50
  Thus Sn=n/2(2a+(n-1)d)
  so Sn=50/2(2+98)=25*100=2500
  2nd series::
  -2,-4,-6,.........-98
  AP with d=2,a=-2 and n=49 from (An=a+(n-1)d)
  So Sn=49/2(-4+(48*-2)=49/2(-100)=49*-50=-2450
  HENCE Sum of both the series gives the ANSWER:::2500-2450=50
• 11 years agoHelpfull: Yes(2) No(0)
• (1+3+........+99)-(2+4+.....+98)
  apply A.P on both we get 2500-2450=50
• 11 years agoHelpfull: Yes(2) No(0)
• Answer is 50.
• 11 years agoHelpfull: Yes(2) No(1)
• 1-2+3-4+......97-98+99
  -1+-1+.....+99= -49+99=50
• 11 years agoHelpfull: Yes(2) No(0)
• odd terms are +VE and even are -VE.
  1+99=100
  -2-98=-100
  .
  .
  .
  47+53=100
  -48-52=-100
  upto these all nullify 100-100=0
  ans is=> 49+51-50=50
  
• 11 years agoHelpfull: Yes(1) No(0)
• i think first 6 digit sum= -3. then -3+7=+4. similarly ..........+99=+50 ans.
• 11 years agoHelpfull: Yes(0) No(0)
• we can write this like,
  (1+3+5+7........99)-(2+4+6+....98)
  (2500)-(2450)-50
• 11 years agoHelpfull: Yes(0) No(0)