Final answer:
Option A: After passing through n blanks, a bullet that loses 1/n of its velocity with each blank will be left with a negligible amount of its initial velocity. The least number of blanks required for the bullet to stop is n.
Step-by-step explanation:
The subject in question involves a mathematical sequence or progression concept where a bullet loses a specific fraction of its velocity after passing through a blank. We can use arithmetic or geometric series to understand how the bullet's velocity decreases with each blank it passes. In the question, a bullet loses 1/n of its velocity each time it passes through a blank. After passing through the first blank, it retains (n-1)/n of its velocity.
Similarly, after passing through the second blank, it retains ((n-1)/n)^2 of its original velocity, and so on. After passing through k blanks, the velocity will be reduced to ((n-1)/n)^k of the initial velocity. In order for the bullet to completely lose its velocity, we need to find the smallest integer k for which this expression is less than or equal to zero. Given that velocity cannot be negative and understanding that fractional velocity will continue to decrease as more blanks are added, the bullet will eventually stop when k reaches n because ((n-1)/n)^n becomes negligible, therefore the least number of blanks required would be n.