Question

There are 8436 steel balls each with a radii of 1 cm stagged in a pile with one ball on top. 3 balls in the second layer, 6 in the third layer, 10 in the fourth layer and so on. Find the number of horizental layers in the pile.

Answer 1

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.