solve
1-2+3-4+5-6........-98+99

• Group the terms as follows:
  1 - 2 + 3 - 4 + 5 - 6 + 7 - ... + 97 - 98 + 99
  = (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + ... + 98).
  
  These are both arithmetic series.
  
  Note that 1 + 3 + 5 + 7 + ... + 99 is an arithmetic series with a first term of 1 and a common difference of 2. It also has 50 terms. Since the sum of an arithmetic series, Sn, with a first term of a and a common difference of d is:
  Sn = (n/2)[2a + (n - 1)d],
  
  we see that:
  1 + 3 + 5 + 7 + ... + 99 = (50/2)[2(1) + (50 - 1)(2)] = 2500.
  
  2 + 4 + 6 + ... + 98 is also an arithmetic series (a = 2, d = 2, n = 49). Thus:
  2 + 4 + 6 + ... + 98 = (49/2)[2(2) + (49 - 1)(2)] = 2450.
  
  Therefore, the required sum is 2500 - 2450 = 50.
  
• 11 years agoHelpfull: Yes(20) No(2)
• Ans: 50
  (1+3+5......99)-(2+4+...........98)
  from AP
  1st series has 50 terms and second has 49 term
  so
  (50*(1+99)/2)-(49*(1+98)/2)
  hence ans would be 50.
• 11 years agoHelpfull: Yes(11) No(1)
• In the whole expression u can see there are 50 terms[(1-2),(3-4)...(97-98) & 99]if we r taking each first 2 digits are a group. So for expression (1-2)+(3-4)+.....+(97-98)= -49 thus ans will be -49+99=50
• 11 years agoHelpfull: Yes(5) No(0)
• 1 - 2 + 3 - 4 + 5 - 6 + 7 - ... + 97 - 98 + 99
  = (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + ... + 98).
  
  These are both arithmetic series.
  
  Note that 1 + 3 + 5 + 7 + ... + 99 is an arithmetic series with a first term of 1 and a common difference of 2. It also has 50 terms. Since the sum of an arithmetic series, S_n, with a first term of a and a common difference of d is:
  S_n = (n/2)[2a + (n - 1)d],
  
  we see that:
  1 + 3 + 5 + 7 + ... + 99 = (50/2)[2(1) + (50 - 1)(2)] = 2500.
  
  2 + 4 + 6 + ... + 98 is also an arithmetic series (a = 2, d = 2, n = 49). Thus:
  2 + 4 + 6 + ... + 98 = (49/2)[2(2) + (49 - 1)(2)] = 2450.
  
  Therefore, the required sum is 2500 - 2450 = 50.
• 11 years agoHelpfull: Yes(4) No(1)
• odd series: n=50 ->n^2=50^2=2500;
  even numbers=49->n(n+1)=49*50=2450;
  odd-even=2500-2450=50;
  answer:50
• 11 years agoHelpfull: Yes(3) No(0)
• ans:50
  (1+3+5.....+99)-(2+4+......+98)=(2500)-(2450)=50
  concept: sum of AP Series...
• 11 years agoHelpfull: Yes(2) No(1)
• ANS- 50
  as per AP series..
• 11 years agoHelpfull: Yes(2) No(1)
• (1+2+3+4+5+6.....+99) + (-4,-8,-12......-196)
  
  4950-4900 = 50 Ans.
• 11 years agoHelpfull: Yes(2) No(0)
• 50 as sum of 1st series=2500 using formua sn=n/2(2a+(n-1)d) and of 2nd series=2450
  then diff bet them is 50
• 11 years agoHelpfull: Yes(1) No(0)
• Answer is 50
  
• 11 years agoHelpfull: Yes(1) No(0)
• 1-2=-1
  3-4=-1
  ......
  ......
  ......
  97-98=-1
  so, 49*-1+99=50
• 11 years agoHelpfull: Yes(1) No(1)
• (1+3+5+7.....99)-(2+4+6......98)
  2500-2450
  (sum of AP starting from 1 to 99 with diff 2)-(sum of AP starting from 2 to 98 with diff 2)
  ans 50
• 11 years agoHelpfull: Yes(0) No(0)
• 1-2+3-4+5-6...........-98+99
  1-2=-1
  3-4=-1
  ....
  97-98=-1
  98/2=49
  -49+99=50
  ans=50
• 8 years agoHelpfull: Yes(0) No(0)