solution: It is a geometric progression. and most appropriate relationship is x×
.
explanation:
let the number of competitors in first round be x.
number of competitors in 2nd round will be
.
number of competitors in 3rd round will be
= ![\frac{x}{4}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B4%7D)
similarly, number of competitors in next round will be
= ![\frac{x}{8}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B8%7D)
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} ,...](https://tex.z-dn.net/?f=x%2C%20%5Cfrac%7Bx%7D%7B2%7D%20%2C%5Cfrac%7Bx%7D%7B4%7D%20%2C%5Cfrac%7Bx%7D%7B8%7D%20%2C%5Cfrac%7Bx%7D%7B16%7D%20%2C...)
now ratio of second and first term =
= ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
similarly, ratio of third and fourth term =
= ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
and so on.
so, it is forming a geometric progression .
where the first term i.e.,a = x
and common ratio i.e., r =
.
so here, most appropriate relationship is ![ar^{n-1}](https://tex.z-dn.net/?f=ar%5E%7Bn-1%7D)
i.e., x×
.