There are N Soldiers in a platoon. They Practice Rifle shooting everyday one by one. A soldier keeps shooting until he hits the target. One day unusual thing happened. First Soldier hits the target in first shot. 2nd soldier hit target in 3rd shots(which is 1+2), 3rd Soldier hit the target in 6th Shot (which is 1+2+3 ) and so on. if Total of 9880 bullets were there, then how many Soldiers were there in Platoon ?

• ans: 38 soldiers
  
  tn = n(n+1)/2
  
  Sum = ∑tn = 1/2 *∑(n^2+n) = 1/2 *[n(n+1)(2n+1)/6 + n(n+1)/2]
  gives
  => n(n+1)(n+2)/6 = 9880
  => n(n+1)(n+2) = 9880*6 = 38*39*40
  => n = 38
  
• 10 years agoHelpfull: Yes(46) No(3)
• Here we have to find sum of 1+3+6+10+15+21+28+........series
  The series is of the form :- n(n+1)/2
  check if n=1 Then ->1
  n=2 ->3
  n=3 ->6
  n=4 ->10 etc
  
  sum of n(n+1)/2 = n^2+n/2---------(1)
  sum of n^2 series is n(n+1)(2n+1)/6
  sum of n term is n(n+1)/2
  on keeping in equation 1 we get
  n(n+1)(2n+4)/12
  
  acc. to question n(n+1)(2n+4)/12 = 9880
  so n= 38..
  you can take value of n from options and can check very easily
• 10 years agoHelpfull: Yes(15) No(1)
• @Rakesh Nice Explanation. and it was asked on 17 august.
• 10 years agoHelpfull: Yes(1) No(0)
• suppose total number of soldier are N then
  total number of shot taken by soldiers=
  (1)+(1+2)+(1+2+3)+.......(1+2+3+...N)=9880
  1/2[1*2+2*3+2*6+..........2(N(N+1)/2))=9880
  
  1/2[(1+1^2)+(2+2^2)+(3+3^2)+......]=9880
  1/2*sumof(N^2+N)=9880
  1/2*[N(N+1)(2N+1)/6+N(N+1)/2]=9880
  
  after simplification
  N(N+1)(N+2)/6=9880
  N(N+1)(N+2)=9880*6
  N(N+1)(N+2)=38*39*40
  
  there for: N=38
  
  
• 10 years agoHelpfull: Yes(1) No(0)
• can anyone post after simplification how it comes n(n+1)(n+2)/6?
• 10 years agoHelpfull: Yes(0) No(0)
• It is given that
  1st Soldier=1
  2nd Soldier=1+2=3
  3rd Soldier=1+2+3=6
  4th Soldier=1+2+3+4=10.....
  and so on using series formula me can solve it
  n(n+1)2=9880
  n^2+n=19760
  now solce this you will get the answer
• 10 years agoHelpfull: Yes(0) No(4)