Question

At a chess tournament, the number of competitors in each round is 50% of the number of competitors in the previous round. What type of relationship most appropriately models this situation?

Answer 1

solution: It is a geometric progression. and most appropriate relationship is x×$(\frac{1}{2} )^{n-1}$.

explanation:

let the number of competitors in first round be x.

number of competitors in 2nd round will be $\frac{x}{2}$.

number of competitors in 3rd round will be $\frac{(\frac{x}{2} )}{2}$ = $\frac{x}{4}$

similarly, number of competitors in next round will be $\frac{(\frac{x}{4} )}{2}$ =  $\frac{x}{8}$

and so on.

so number of competitors in various rounds are forming a sequence

i.e., $x, \frac{x}{2} ,\frac{x}{4} ,\frac{x}{8} ,\frac{x}{16} ,...$

now ratio of second and first term =  $\frac{(\frac{x}{2} )}{x}$ =  $\frac{1}{2}$

similarly, ratio of third and fourth term =  $\frac{(\frac{x}{4} )}{\frac{x}{2}}$ =  $\frac{1}{2}$

and so on.

so, it is forming a geometric progression .

where the first term i.e.,a = x

and common ratio i.e., r = $\frac{1}{2}$.

so here, most appropriate relationship is $ar^{n-1}$

i.e., x×$(\frac{1}{2} )^{n-1}$.