The 20 th term from the end of the progression 20 , 19 1 4 , 18 1 2 , 17 3 4 , … , - 129 1 4 is :-

Given $AP$ is $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},...,-129\frac{1}{4}$.

Common difference, ${\mathrm{d}}_1=19\frac{1}{4}-20=-\frac{3}{4}$

Reversing the $AP$ such that first term becomes the last and last term becomes first.

$\Rightarrow -129\frac{1}{4},......,19\frac{1}{4},20$

This is also $AP$.

First term, $\mathrm{a}=-129\frac{1}{4}$ and common difference, $\mathrm{d}=\frac{3}{4}$.

$\Rightarrow a_{20}=-129\frac{1}{4}+(20-1)\left(\frac{3}{4}\right)$

$\Rightarrow a_{20}=\frac{-517}{4}+\frac{57}{4}$

$\Rightarrow a_{20}=\frac{-460}{4}$

$\Rightarrow a_{20}=-115$