The 20 ^ th term from the end of the progression 20, 19(1)/(4),...

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

Common difference, ${d}_{1} = 19\dfrac{1}{4} - 20 = - \dfrac{3}{4}$

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

$⇒ - 129\dfrac{1}{4}, . . . . . . , 19\dfrac{1}{4}, 20$

This is also $AP$.

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

$⇒ {a}_{20} = - 129\dfrac{1}{4} + ( 20 - 1 ) \left(\dfrac{3}{4}\right)$

$⇒ {a}_{20} = \dfrac{- 517}{4} + \dfrac{57}{4}$

$⇒ {a}_{20} = \dfrac{- 460}{4}$

$⇒ {a}_{20} = - 115$