if n is a natural no,
101+102+....n what is the probability that this sum is divisible bu (n-100)?

• probability =50%
  coz..101/1, 306/3, 515/5, 728/7, 945/9..n so on
  so odd nos are divisible...while even nos are not divisible giving odd sum
• 11 years agoHelpfull: Yes(24) No(2)
• 1/2
  since take add terms like 101(div by 1(101-100)),3rd term(upto 103(306/3).....
  even terms (101+102) not div by 2 ..so sum is div wen it is odd ..and not div n is even
• 11 years agoHelpfull: Yes(4) No(3)
• Ans is 50%
  cz 101/(101-100)=101, (101+102)/(102-100)=203/2, (101+102+103)/(103-100)=102, (101+102+103+104)/(104-100)=410/4..
  from the abve it clear tht odd nos r divisible and even nos r nt divisible.. so result 50%.
  
• 11 years agoHelpfull: Yes(2) No(0)
• visit this for solution http://www.onlineadhyayana.com/ elitmus section...:)
• 11 years agoHelpfull: Yes(1) No(2)
• hi pallavi, how u done d division for 101/1,306/3,525/5,728/7....these numbers? 101,306,515,728...wat are these numbers.cann u xplain me in detail
• 11 years agoHelpfull: Yes(1) No(1)
• and in ques they also given (n>100)
• 11 years agoHelpfull: Yes(0) No(0)
• i hav given the explanation but it onli some part is visible ...
  @praveen is ma ans corret?
• 11 years agoHelpfull: Yes(0) No(0)
• i hav given the explanation but it onli some part is visible ...
  @praveen is ma ans correct?
• 11 years agoHelpfull: Yes(0) No(0)
• sum of this series would be (n^2+n-10100)/2
  clearly it is divisible by (n-100)
  hence probability must be 1
  
• 11 years agoHelpfull: Yes(0) No(0)
• jjljk hh jl
• 11 years agoHelpfull: Yes(0) No(0)
• probability = 1/2
  number of terms : n-101+1=n-100
  Sum of numbers using AP formula : (n-100)*(n+101)/2
  Sum will be divisible by (n-100) only when (n+101) is divisible by 2
  This is possible only when n is odd. [ coz odd number added with odd number gives even number]
  Probability of n being odd = 1/2 [Since, only two possibilities are there, n is even or n is odd]
• 9 years agoHelpfull: Yes(0) No(0)