what is the sum of all even integers between 99 and 301?
a)40000
b)20000
c)40200
d)20200

• Ans is 20,200
  as,we have sequence from 100,102,...,288,300.
  so a=100,d=2,tn=300,so we have 101 elements,as per formula tn=a+(n-1)d.
  and now sum=n/2(a+l)=101/2(100+300)=20200
• 12 years agoHelpfull: Yes(71) No(8)
• 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
  
  anthor method seqence is an A.P with first tem 100, common difference 2, no.of terms =101
  
  sum = 101*(100+300)/2=20200
• 12 years agoHelpfull: Yes(33) No(3)
• So if the smallest and largest integers are 100 and 300,the mean is 200
  Next, the number of terms. Between 100 and 199, inclusive, for instance,
  there are exactly 100 integers,50 of which are even. If we look at 100 to 200,
  inclusive, there are 51 evens. By the same principle, between 100 and 299,
  inclusive, there are exactly 200 integers, 100 of which are even. Include 300,
  and there are 101.
  Thus, there are 101 terms
  sum, then, is:
  200(101) = 20200(ans)
• 12 years agoHelpfull: Yes(16) No(9)
• here 99 to 101 is given as we are finding sum of even no so we will take 100 to 300.
  Now,
  n=300-100/2 + 1 =101 no of even no are there.
  Now to get the sum of series:
  
  Required sum=[n/2](a+l)
  a=100, b(step difference)=2, l=300
  So, [101/2](100+300)=(400*101)/2 =40400/2 =20200
• 12 years agoHelpfull: Yes(13) No(3)
• ANSWER IS d) 20200
  since the nos start from 100 to 300
  nw n=300-100/2 + 1 =101
  use sum of n terms frmula of a.p
  u will get d)
• 12 years agoHelpfull: Yes(8) No(3)
• Formula for sum of the arithmetic series = n/2[2*a + (n-1)*d]
  and the even numbers between 99 to 301 is 100 to 300
  between 100 to 199 =>50numbers
  between 200 to 299 => 50 numbers
  and one is 300
  so total even numbers between this range is = 50+50+1 => 101
  n=101(total numbers)
  a=200(first number)
  d=2(difference between two consecutive number of this series)
  Put in the formula : 101/2[2*200+(100)*2]
  (101/2)*400
  101*200 = 20200
• 12 years agoHelpfull: Yes(6) No(5)
• Even Numbers between 99 and 301 is 100,102,104,106,108.......298,300.
  this is in the form of arithmetic progression
  first number,A=100 Last number,L=300
  total number of even numbers between 99 & 301 is N = 101.
  Formula for total sum=0.5*N*(A+L) = 0.5 * 101 * (100+300) = 20,200
  
  
• 11 years agoHelpfull: Yes(2) No(0)
• 100 to 300
  100+102+104....+300= 2*(50+51+52....150)
  sum of 50 to 150 integers is = sum of 0 to 150 minus sum of 0 to 49
  i.e, (150*151/2)-(49*50/2)=10100
  so 10100*2=20200
  
• 10 years agoHelpfull: Yes(2) No(0)
• the even integers is:100,102,104,---------300
  by using the formula n/2[a+l]
  
  now claculate the n:
  a+(n-1)d=300
  100+(n-1)2=300
  100+2n-2=300,n=101
  
  sum=n/2[a+l]
  =101/2[100+300]
  =20200
• 10 years agoHelpfull: Yes(1) No(0)
• a=100, tn=300,d=2
  and tn=a=(n-1)d
  => 300=100+(n-1)*2
  =>n=101
  
  and now sn=n/2(a+tn)=20200
• 9 years agoHelpfull: Yes(1) No(0)
• a=100
  d=2
  l=300
  n=101
  sum=n(a+l)/2
  sum=101*400/2
  sum=20200
• 9 years agoHelpfull: Yes(1) No(0)
• first term=100
  second term=300
  number of difference=2
  so sum of all even integers= 100/2(100+300) =50*400=20,000
• 9 years agoHelpfull: Yes(1) No(1)
• sum of first n numbers is given by n(n+1)/2
  Even numbers from 99 to 301 is 100,102,104...300
  i.e 2(50,51,52...150)=total 101 terms.
  to get answer first find sum from 1-49 and second find sum from 50-150
  then subtract first sum from second sum u get the answer...Multiply the second sum with 2 before subtracting.
  u will get 20200
• 10 years agoHelpfull: Yes(0) No(0)
• 1st term 100,2nd term=300
  the difference is 2.
  so
  (300-100)/2=100.
  
  sum of all even number between 99 and 301 is
  
  100/2*(100+300)=20000.
• 10 years agoHelpfull: Yes(0) No(0)
• 100+102+104+106........300
  Formula n(a+l)/2 where n is no of terms . a is first term and l is last term.
  Here n can be determined by formula n= a+(n-1)d
  Final ans is 101*(100+300)/2
  101*200=20200
• 9 years agoHelpfull: Yes(0) No(0)
• frnds its easy
  
  sum of first n even natural number is n(n+1)
  till 99 there are 49 even integers are there
  till 301 there are 150 integers.
  now put these values one by one in the formula
  and substract 2nd equation from first
  that vl be the answer
  i vl show u
  (150*151)-(49*50)=20200
  thats the easiest way i think
• 9 years agoHelpfull: Yes(0) No(1)
• 20200 is the answer
• 5 years agoHelpfull: Yes(0) No(0)