3. Soldier were practicing to shoot. Coincidence occurred, first soldier shoot the target in his first chance, second shoot the target in his 3rd chance (i.e. 1 +2), third shoot in his 6th chances (i.e. 1+2+3), and so on.. Total 9880 bullet were fired. Find the number of soldier.

• series formed will be
  1 3 6 .....
  it's in the form of n(n+1)/2
  
  sum of ( n(n+1)/2) = 9880
  1/2 sum of (n^2+n)=9880
  1/2 [ n(n+1)(2n+1)/6 + n(n+1)] = 9880
  n(n+1)(n+2) = 59280
  38*39*40 = 59280
  
  therefore n= 38 = no. of soldier
• 10 years agoHelpfull: Yes(17) No(2)
• 1st soldier shoot = 1 chance
  2nd soldier shoot = 1+2 chance
  3rd soldier shoot = 1+2+3 chance
  4
  .
  .
  .
  nth soldier shoot = 1+2+3+4+.......+n chance
  so,
  1+2+3+4+.......+n= n*(n+1)/2
  Total 9880 bullet were fired, so
  n*(n+1)/2 =< 9880
  solve hit and trail method
  when n=140 then total bullet fired 9800
  the 141 men are firing other(80) bullet but not get shoot.
  so,
  141 soldier........ ANS.
• 10 years agoHelpfull: Yes(3) No(3)
• the series formed is- 1+3+6+10+15....
  it is in the form of 1/2*(n(n+1))
  summation of 1/2*(n(n+1))= 9880
  i.e summation of 1/2*(n^2+n)=9880
  1/2[n(n+1)(2n+1)/6+n(n+1)/2]=9880
  n*(n+1)*(2n+4)=118560
  on checking, 38*39*80=118560
  so, n=38
• 10 years agoHelpfull: Yes(1) No(1)
• plz xpln 5th step how is it??
• 10 years agoHelpfull: Yes(1) No(0)
• take 1+3+6+10+15+21 nw this will be the series and sum upto
  1+3+6+10+15+21...=9880
  n/2*(n+1) =9880
  now take summation of above equation by solving it we get a quadtatic equation solve it the ans will be 40
• 10 years agoHelpfull: Yes(0) No(3)
• sry frnds for mis typing
  when n=140 then total bullet fired 9800 (wrong)
  ....................................9870
  
• 10 years agoHelpfull: Yes(0) No(3)