There are 240 balls and n number of boxes B1, B2, B3, … , Bn. The balls are to be placed in the boxes such that should contain 4 balls more than B2, B2 should contain: 4 balls more than B3, and so on.

Explanation

The problem involves distributing balls into a series of boxes so that each box has 4 balls more than the next one. This creates an arithmetic series. The sum of an arithmetic series can be given by the formula: n/2(2a+(n-1)d), where `n` is the number of terms (boxes), `a` is the first term, and `d` is the common difference. Given we have 240 balls, the number of balls in the first box is 4(n-1) since each box has 4 balls more than the next one. Applying the sum formula, we get 240 = n/2 [2*4(n-1) + (n-1)4] 240 = 2n^2 – 2n Solving this quadratic equation, we find no integer solutions for n=7. The answer options `4`, `5`, `6` yield integer solutions satisfying the condition, but `7` doesn`t. Hence, the number of boxes cannot be 7 i.e., option-4.