The 20th term from the end of the progression $$20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \ldots, -129\frac{1}{4}$$ is :

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

These terms decrease by a fixed amount, so it is an arithmetic progression (A.P.).
First term, $$a = 20$$.
Common difference, $$d = 19\frac{1}{4} - 20 = -\frac{3}{4}$$.
Last term, $$l = -129\frac{1}{4}$$.
For an arithmetic progression with last term $$l$$ and common difference $$d$$, the $$k$$-th term from the end is
$$T_{\text{end},\,k}= l - (k-1)d \quad -(1)$$
Here $$k = 20$$, so substitute in $$(1)$$:
$$T_{\text{end},\,20}= l - 19d$$
$$T_{\text{end},\,20}= -129\frac{1}{4} - 19\left(-\frac{3}{4}\right)$$
$$= -129\frac{1}{4} + 19\cdot\frac{3}{4}$$.
$$19\cdot\frac{3}{4}= \frac{57}{4}=14\frac{1}{4}$$.
$$T_{\text{end},\,20}= -129\frac{1}{4}+14\frac{1}{4}= -115$$.
The 20-th term from the end is $$-115$$, which matches Option C.