The 20^th term from the end of the progression $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots <!--\; \dots \; -->.,-<!--\; -\; -->129\frac{1}{4}$ is :

The correct option is -115.

To find the 20^th term from the end of the given progression, we first identify the type of progression and its key components.

The given progression is:
$$20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots ,-129\frac{1}{4}$$

Let's convert the mixed fractions into improper fractions to make calculations easier:
First term ($a_1$) = $20$
Second term ($a_2$) = $19\frac{1}{4}=\frac{77}{4}$
Third term ($a_3$) = $18\frac{1}{2}=\frac{37}{2}=\frac{74}{4}$
Last term ($l$) = $-129\frac{1}{4}=-\frac{517}{4}$

This is an Arithmetic Progression (AP) because the difference between consecutive terms is constant:
Common difference ($d$) = $a_2-a_1=\frac{77}{4}-20=\frac{77-80}{4}=-\frac{3}{4}$

The formula to find the $n^{\mathrm{th}}$ term from the end of an Arithmetic Progression is given by:
$$T_{n\text{from end}}=l-(n-1)d$$

Here, we want to find the 20^th term from the end ($n=20$). Substituting the values of $l=-\frac{517}{4}$ and $d=-\frac{3}{4}$ into the formula:
$$T_{20}=-\frac{517}{4}-(20-1)(-\frac{3}{4})$$

Simplifying the expression step-by-step:
$$T_{20}=-\frac{517}{4}-19(-\frac{3}{4})$$
$$T_{20}=-\frac{517}{4}+\frac{57}{4}$$
$$T_{20}=\frac{-517+57}{4}$$
$$T_{20}=\frac{-460}{4}$$
$$T_{20}=-115$$

Thus, the 20^th term from the end of the progression is indeed -115.