Wat is d sum of all even integer between 99 and 301.
a. 40000
b. 20000
c. 40400
d.20200

• 100,102,104, ...296,298,300
  
  100+300=400
  102+298=400
  104+296=400
  .
  .
  198+202=400
  there 50 terms of sum=400 and 200
  
  answer is 50*400+200=20200
• 12 years agoHelpfull: Yes(24) No(0)
• 100,102,104, ...296,298,300
  these are the even number ...
  let there be n number of terms
  t(n)=1st term + (n-1)d
  300=100+(n-1)2
  n=101
  so sum=n(a+l)/2 where l=last term. a= 1st term.
  sum=101*(100+300)/2=20200
  
• 12 years agoHelpfull: Yes(20) No(1)
• t(n)=a+(n-1)d
  300=100+(n-1)2
  n=101
  sum=101*(100+300)/2=20200
• 12 years agoHelpfull: Yes(13) No(0)
• no. of terms =202
  so formula is 202*(99+301)/2=40400 (ans)...
• 12 years agoHelpfull: Yes(6) No(10)
• ans:20200
  the first even num is:102
  the last even num is :298
  sum===400
  the from 99 to 301 we got 50 even num
  400*50=20000
  the remaing 200 is left adding the one tha is=20000+200
  20200
• 12 years agoHelpfull: Yes(3) No(0)
• ANS IS D 20200
• 12 years agoHelpfull: Yes(2) No(0)
• SRAVYA.. thnks fr d ansr it is correct
  
• 12 years agoHelpfull: Yes(0) No(0)
• 20200...
  starting from 100 to 300.. total 101 terms.
  use formula for the sum of AP
• 12 years agoHelpfull: Yes(0) No(0)
• Answer is D
  1st even no. will be 100 and the last will be 300 with a common difference of 2.
  So the no of terms between 100 and 300 that are even can be fount out by
  300=100+(n-1)*2 that is equal to 101.
  Now applying S=n(a+l)/2 which will give S=20200
  
• 12 years agoHelpfull: Yes(0) No(0)