What is the sum of all even integer between 99 and 301?

• even integers betn 99 & 301 are starting from 100 & last is 300
  
  Sum=100+102+104+...+300
  its an AP with a=100,d=2,l=300,n=no. of terms
  l=a+(n-1)*d => 300=100+(n-1)*2 => n=101
  
  Sum = (n/2)*(a+l)= 101/2 *(100+300)= 101*200 = 20200
  
  
• 10 years agoHelpfull: Yes(37) No(1)
• even integers betn 99 & 301 are starting from 100 & last is 300
  
  Sum=100+102+104+...+300
  its an AP with a=200,d=2,l=300,n=no. of terms
  l=a+(n-1)*d => 300=100+(n-1)*2 => n=101
  
  Sum = (n/2)*(a+l)= 101/2 *(100+300)= 101*200 = 20200
• 10 years agoHelpfull: Yes(5) No(0)
• Ans: 20200
  exp:
  it is an ap Ist term is 100 last term =300
  total no of terms={300-100}/2 + 1 =101 { +1 because both 100 and 300 are included }
  
  sum of n terms = no of terms * { first term + last term}/2
  =101*{100+300}/2
  =101*400/2
  =101*200
  ans =20200
• 10 years agoHelpfull: Yes(2) No(0)
• series will be 100,102,104........300
  so total number in the series =[(300-100)/2 + 1] because both the ends are inclusive n=101
  sum of such series =n/2*[a+l]
  =>101/2*[100+300]
  =>101*200
  =20200
• 10 years agoHelpfull: Yes(1) No(0)
• s = n[a+(n-1)*d/2]s
  
  n = no of terms = 150
  a = first term = 100
  d = difference = 2
  
  so the answer is 37350
• 10 years agoHelpfull: Yes(1) No(3)
• Answer: 20200
  2(50+51+52........150)
  2((150*151/2)-(49*50/2))
  22650-245=20200
• 10 years agoHelpfull: Yes(0) No(0)
• is it 300=100+(n-1)*2 0r 300=100+(n-1)^2 ?
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• Even integers between 99 and 301 are 100, 102, 104 ......,298,300
  Their are 101 terms.
  Their Sum =100+102+104+...+300
  This an AP with first term a=100,common difference d =2 and last term l=300
  Using formula Sum = (n/2)*(a+l)
  We get Sum = (101/2)*(100+300)= 101*200 = 20200
  
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• even integers between 99 and 301 starts from 100 and ends at 300.
  now [(100+300)/2]*[[(300-100)]/2+1]
  =200*101=20200
• 10 years agoHelpfull: Yes(0) No(0)
• 1st & last even integer between 99 & 301 are 100 & 300 respectively. so we can say that the nth term 300=100+(n-1)2(as nth term=a+(n-1)d, where a=1st term of series, n=no. of terms & d=common difference).
  so from the above eqution we get n=101.
  now sum of series=n/2[2a+(n-1)d].
  sum of even integer between 99 and 301=101/2[2*100+(101-1)2]=20200
• 10 years agoHelpfull: Yes(0) No(0)