Find sum of the series 1-2+3-4+....-98+99

• ans z 50.v can write as-
  (1+3+.....+99)-(2+4+....+98)ie 50 terms+49 terms resp.
  using formla s=n/2[a+l]
  we get, 50/2[1+99]-49/2[2+98]
  den 2500-2450=50.
• 10 years agoHelpfull: Yes(19) No(2)
• its simple,
  lets take first 5 no's,
  1-2+3-4+5=3=(middle no. or (5+1)/2=3)
  lets take first 7 no's,
  1-2+3-4+5-6+7=4=(middle no. or (7+1)/2=4)
  lets take first 11 no's,
  1-2+3-4+5-6+7-8+9-10+11=6=(middle no. or (11+1)/2=6)
  so,
  1-2+3-4+5-6+....-98+99 = (99+1)/2 = 50
  ans:- 50
  
• 10 years agoHelpfull: Yes(10) No(0)
• (1-2)+(3-4)+(5-6)+.....+(97-98)+99
  (-1)+(-1)+......(-1)+99
  -49+99
  50
  
• 8 years agoHelpfull: Yes(4) No(0)
• answer is -49+99=50
• 10 years agoHelpfull: Yes(2) No(3)
• Ans is = 99
  sum of odd series = 1+3+5+...+99 = n^2=99^2
  sum of even series = 2+4+6+...+98= n(n+1)=98(99)
  so sum = 99^2-98(99)=99
• 10 years agoHelpfull: Yes(1) No(3)
• ans.1
  (1+3+5+...)-(2+4+..)
  25(2+49)-49(4+48)/2
  =1275-1274=1
• 10 years agoHelpfull: Yes(0) No(1)
• by n/2(a+l)
  25(1+99)-49/2(2+98)
  25*100-49*50=2500-2450
  =50
• 9 years agoHelpfull: Yes(0) No(0)
• (1-2) + (3-4) ........ + (97-98) = -1 + -1 ......... + -1 like that pairing upto -98 = -49
  then -49 + 99 = 50
• 7 years agoHelpfull: Yes(0) No(0)
• Ans is -50
  1+99 = 100
  -2+(-98) = -100
  3+97=100
  -4+(-96)=-100
  
  similarly,
  49+51=100
  
  hence ans is -50, because all even numbers are negative as per series.
• 6 years agoHelpfull: Yes(0) No(0)