Q. A group of 630 children arranged in group.each row contain 3 fewer children than the row front of it.how many rows of numbers are possible?

• Answer: 21
  Explanation:Each row contains 3 greater than the front row..
  So, the order follows...
  1st row 3, 2nd row 6 and so on.... which indicates an Arithemetic Progression(A.P)
  we don't know no.of terms in the series... but the sum is 630 i.e., the total no.of children.
  n(2a+(n-1)d)/2 = 630
  a=3, d=3
  substituting and solving we get n=21,the no.of rows required....
• 11 years agoHelpfull: Yes(3) No(1)
• number of rows will be 20.
  check
  
  n(2a+(n-1)d)/2 = 20(6+19*3)/2= 10*63=630
• 11 years agoHelpfull: Yes(1) No(0)
• Answer is 20
  
  3(1+2+3+.....N)=630
  3[N(N+1)/2]=630
  N(N+1)/2=210
  N(N+1)=420
  solving we get N=20,-21
  so No. of possible rows are 20.
  
• 11 years agoHelpfull: Yes(0) No(0)
• i dunno d ans.
  
• 11 years agoHelpfull: Yes(0) No(0)