What is the sum of all even integers between 99 and 301?
Option
A. 40000
B. 20000
C. 40400
D. 20200

• D. 20200
  Even integers between 99 and 301 are 100,102,104,....,300
  To find total even numbers n , 300=100+(n-1)2, n=101
  Sum of all numbers=(101/2)[2*100+ (101-1)2]
  =10100+10100 =20200
• 10 years agoHelpfull: Yes(28) No(0)
• 20200
  
  100+102+104+....298+300 = (2+4+6+...298+300) - (2+4+6+...+96+98)
  =2*(1+2+3+...+150) - 2*(1+2+3+...49)
  =150*151 - 49*50
  =20200
• 10 years agoHelpfull: Yes(9) No(1)
• ACCORDING TO A.P
  A=100 L=300 n=? D=2
  300=100+(n-1)2 (Tn=a+(n-1)d)
  200=(n-1)2
  n=101
  
  so, Sn=(101/2)*(100+300) (Sn=(n/2)*(a+l))
  =40400/2
  =20200
  answer= 20200
• 10 years agoHelpfull: Yes(7) No(1)
• ans:D
  frst & last even nums between 99 and 301 are 100 & 300 respectivly..
  in Arthmtic progression,
  no.of even integers n= [(last num- frst term)/commn diffrnce]+1
  n=[(300-100)/2]+1=101
  sum=n/2[frst term+last term]
  sum=101/2[100+300]=20200
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• yes D. is the rght ans
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• 100+102+104+..........+300
  tn=300,a=100,d=2
  tn=a+(n=1)d,solving it we get n=101
  sn=n/2(2a+(n-1)d),solving it ans is 20200
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• sum = 100+102+........+300
  it is an A.P where d=2 , n=101[(300-100)/2 + 1 =101]
  sum=101/2[2x100 + (101-1)x2]
  =20200
  option d
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• n/2[a+l] formula n=101 a=100 l=300
  ans 20200
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• total elements are 100
  1st term is 100 and last term is 300
  then the total sum of the even elements is
  100(100+300)/2=20000
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• ans is 20200
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• ans is 20200
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• 300=100+(n-1)2
  200/2=n-1
  101=n
  S=((100+300)*101)/2
  s=20200
  so option d is correct
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• answer id D
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