find the value of the expression
1-2+3-4+5-6+.......-96+97-98+99

• (1+3+5...99)-(2+4+...98)
  let it be n1-n2 where n1=1+3+5+7...99 & n2=2+4+6...+98
  n1 & n2 are in A.P. so we will solve it seperately
  In case of n1
  ...................
  first term=1
  last term =99
  so,99=1+(n-1)2
  n=50
  so n1=sum=50(1+99)/2=2500
  In case of n2
  ..................
  first term=2
  last term=98
  so,98=2+(n-1)2
  n=49
  so sum=n2=49(2+98)/2=2450
  so,n1-n2=2500-2450=50
• 10 years agoHelpfull: Yes(44) No(1)
• 50
  1-2+3-4+.....-96+97-98+99
  total no are 99
  -1-1-1(49 times)+99=50
• 10 years agoHelpfull: Yes(26) No(0)
• we know sum of first n odd numbers=n*n,first even numbers=n*(n+1),
  then my answer is
  1+3+.....+99=50*50
  2+4+......+98=49(49+1)
  now answer is=(50*50)-(49*50)=50.its simple
  
• 10 years agoHelpfull: Yes(12) No(3)
• we get 49 pairs of -1..like 1-2=-1 then 3-4=-1 so last 99-49=50
  
• 10 years agoHelpfull: Yes(9) No(2)
• (1+3+5+.....+99)-(2+4+6+....+98)
  first we calculate the no of terms in first A.P. series
  99=1+(n-1)*2
  (99-1)/2=(n-1)
  n=49+1=50
  now we calculate the no of terms in second A.P. series
  98=2+(n-1)*2
  (98-2)/2=(n-1)
  n=48+1=49
  (1+3+5+...+99)-(2+4+6+...+98)
  ((n/2)*(a+l))-((n/2)*(a+l))
  ((50/2)*(1+99))-(49/2)*(2+98)
  (25*100)-(50*49)
  2500-2450=50
• 10 years agoHelpfull: Yes(2) No(1)
• 50
  (1+3+.....99)-(2+4+....98) both the series are in A.P.
  using summation formula...
  sum of odd terms=2500
  sum of even terms=2450
  ans=2500-2450=50
• 10 years agoHelpfull: Yes(2) No(0)
• 1+3+.....+99= 50*50(total no present in sum)^2;
  2+4+.....+98=49*50(total no present in sum)*(total no present in sum+1)
  =50*50-(49*50)
  =50(50-49)
  =50
  
• 10 years agoHelpfull: Yes(1) No(2)
• it is the sum of odd numbers-sum of even numbers
  sum of odd numbers=n/2+n/2=(n^2)/4=(10000/4)=2500
  sum of even numbers=n/2*(n/2+1)=50*51=2550
  The difference is odd-even ...ans:2550-2500=50
• 10 years agoHelpfull: Yes(1) No(0)
• (1+3+5+......+.....+99)-(2+4+6+8+........98)
  let us find the sum of AP of given series.
  1+3+5+7+....99 2nd series =(2+4+6+....98);
  =2450
  99=1+(n-1)*2
  n=50.
  now sn = n/2*(a+l);
  = 50/2*(1+99);
  =2500;
  now above eqn.
  = 2500-2450
  =50ans.
• 10 years agoHelpfull: Yes(1) No(0)
• 1-2=-1
  3-4=-1
  97-98=-1
  so total we have 98/2=49 -1's
  so
  -49+99=50
• 10 years agoHelpfull: Yes(1) No(1)
• 1-2 gives -1
  3-4 gives -1 and so on
  therefore we have -1-1-1.........49times+99= -49+99=50
• 10 years agoHelpfull: Yes(1) No(0)
• 99 as it is an odd number so (99+1)/2=50 ans.
• 10 years agoHelpfull: Yes(1) No(2)
• from 1 to 98 there are 98 numbers.
  so 98/2 = 49
  1-2+3-4+5-6+.......-96+97-98+99
  = -49 + 99
  =50
• 10 years agoHelpfull: Yes(1) No(0)
• 50
  The given expression can be written as
  = (1+3+5+....+99)-(2+4+6+....+98)
  
  = 50/2(100) - 49/2(100)
  
  = 50(50-49)
  
  = 50
  
• 10 years agoHelpfull: Yes(0) No(1)
• answer=50;
  sum of all odd term from 1 to 99 - sum of all even term 2 to 98;
  apply simple A.P. series to get sum ,TOTAL ODD TERM ARE 50 &EVEN TERM ARE 49;
  =50*50-50*49=50;
• 10 years agoHelpfull: Yes(0) No(0)
• (1+3+5+...+99)-(2+4+6+...+98)
  =2500-2450
  =50
• 10 years agoHelpfull: Yes(0) No(0)
• for odd numbers : 1+99 =100 ; 3+97 =100 ; 5+95=100 ...so on
  for even numbers : -2 -98 = -100 ; -4 - 96 = -100 ; -6 -94 = -100 ...so on
  hence in the middle there will be only 49 + 51 =100 and -50 implies 100-50 = 50
• 10 years agoHelpfull: Yes(0) No(0)
• sum of odd numbers is 2500
  sum of even numbers is 2450
  2500-2450=50
• 10 years agoHelpfull: Yes(0) No(0)
• 1-2+3-4+.........-98+99
  1+3+5......._(2+4+.....98)
  2500-2450=50
• 10 years agoHelpfull: Yes(0) No(0)
• make two series of A.P by this expression
  such that
  1+3+5.....+97+99-(2+4+6....+96+98)
  50/2(2*1+(50-1)2) - 49/2(2*2+(49-1)2) = 50
• 10 years agoHelpfull: Yes(0) No(0)
• For every two value,we have got -1.for example,1-2=-1,3-4=-1, so we have got
  -49+99=50
• 10 years agoHelpfull: Yes(0) No(0)
• (1+3+5+.....+99)-(2+4+6+....+98)
  n A.P. series
  a=1 and l=99
  99=1+(n-1)*2
  where n=50
  (99-1)/2=(n-1)
  n=49+1=50
  now we calculate the no of terms in second A.P. series
  98=2+(n-1)*2
  (98-2)/2=(n-1)
  n=48+1=49
  (1+3+5+...+99)-(2+4+6+...+98)
  ((n/2)*(a+l))-((n/2)*(a+l))
  ((50/2)*(1+99))-(49/2)*(2+98)
  (25*100)-(50*49)
  2500-2450=50
• 10 years agoHelpfull: Yes(0) No(0)
• Up to 98 there are 98/2 times sum of -1. so sum = -49.
  Last term is 99
  so value of expression = -49 + 99 = 50
  
• 10 years agoHelpfull: Yes(0) No(0)
• 1-2+3-4+5-6+.......-96+97-98+99
  =(1-2)+(3-4)+(5-6)+.....+99
  =-1-1-1.......+99
  now for 1st 6 terms there are 3 no. of -1's
  so for 98 terms there are 98/2=49 no. of -1's
  +99(last term)
  =-49+99
  =50(Ans)
• 10 years agoHelpfull: Yes(0) No(0)
• (1+3+5.....+99) - (2+4+....+98)
  = 2500 - 2450
  =50
• 10 years agoHelpfull: Yes(0) No(0)
• 1+3+.....+99=50*50
  2+4+......+98=49(49+1)
  now answer is=(50*50)-(49*50)=50.
• 10 years agoHelpfull: Yes(0) No(0)
• sum of 50 terms-sum of 49 terms=2500-2450=50
• 10 years agoHelpfull: Yes(0) No(0)
• answer is 0.
• 10 years agoHelpfull: Yes(0) No(4)
• There are 49 sets which results to 1...(-2+3)+(-4+5)+....+(-98+99)+1=49+1=50
• 10 years agoHelpfull: Yes(0) No(0)
• 0 ... Its an a.p of odd minus even terms
• 10 years agoHelpfull: Yes(0) No(2)
• 0
  Taking odd numbers common and even numbers common then expression becomes
  (1+3+5...99)-(2+4+6+98)
  Two sequences are in AP.
  So sum of terms in AP (a+l)/2 if first term is a and last term is l.
  Now applying formula
  (1+99)/2-(2+98)/2=0
  
  
• 10 years agoHelpfull: Yes(0) No(1)
• 1-2=-1
  3-4=-1
  so,from 1 to 98 -49
  so,99-49=50
  ans=50
• 10 years agoHelpfull: Yes(0) No(0)