7880 ball are there of diameter 2 cm, if these ball are arranged in a form of pile such that top has 1 ball,2nd layer has 3,3rd has 6,4th has 10,5 has 10 and so on.how many levels are there.
Options are
a. 34
b. 33
c. 31
d. 32

• its a series of 1,3,6,10,15,21, so on
  here, each term varies in series like +2,+3,+4,+5,+6 so on
  so, sum upto n terms i.e. 1+3+6+10+15+21+... upto n terms =7880
  here after solving this AP, n= 34
  
• 10 years agoHelpfull: Yes(2) No(6)
• are idiot uetion post karne ka man hi hai to thik s post karo naa
• 10 years agoHelpfull: Yes(1) No(2)
• @puja
  
  its not in AP, then how to get the sum ?
• 10 years agoHelpfull: Yes(1) No(0)
• actually the ques is wrong ..ques is
  There are 7880 balls, each with a radius of 2 cm stacked in a pile with 1 ball on top,3 balls in the second layer,6 in the 3rd layer,10 in the fourth and so on, the number of horizontal layers in the pile is ?
• 7 years agoHelpfull: Yes(1) No(0)
• The given series is this: 1, 3, 6, 10, 15....
  Note, every nth term is the sum of first n positive integers i.e. 3rd term is 1+2+3 = 6 and so on.
  
  Therefore t(n) = n(n+1)/2 = (n^2)/2 + n/2
  Sum of n terms of this series will be S(n) = ∑t(n) = (1/2)∑n^2 + (1/2) ∑n (summation of the terms)
  This gives us S(n) = (1/2) n(n+1)(2n+1)/6 +(1/2)n(n+1)/2
  Take n(n+1) common from the two terms and simplify the rest to get S(n) = n(n+1)(n+2)/6
  
  Now n(n+1)(n+2)/6 = 8436
  or n(n+1)(n+2) = 8436 x 6
  At this point, use options. If n = 34, product of 34, 35, 36 will end in 0, not 6, so not possible.
  If n = 36, product of 36,37 and 38 will end in 6. This is the only possibility. None of the other products will end in 6 so answer must be 36.
• 7 years agoHelpfull: Yes(1) No(0)
• we find h.c.f of 7880 so ans is 32.....
• 10 years agoHelpfull: Yes(0) No(3)
• pls explain
  
• 10 years agoHelpfull: Yes(0) No(1)
• this link help you guys
  http://gmatclub.com/forum/sum-of-series-103317.html
• 10 years agoHelpfull: Yes(0) No(0)