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 :

Answer :

Solution :

$20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots <!--\; \dots \; -->\dots <!--\; \dots \; -->.,-<!--\; -\; -->129\frac{1}{4}$

This is A.P. with common difference

${\mathrm{d}}_1=-<!--\; -\; -->1+\frac{1}{4}=-<!--\; -\; -->\frac{3}{4}$ $$

$-<!--\; -\; -->129\frac{1}{4},\dots <!--\; \dots \; -->\dots <!--\; \dots \; -->\dots <!--\; \dots \; -->\dots <!--\; \dots \; -->...19\frac{1}{4},20$

This is also A.P. $\mathrm{a}=-<!--\; -\; -->129\frac{1}{4}$ and $\mathrm{d}=\frac{3}{4}$

Required term $-<!--\; -\; -->129\frac{1}{4}+(20-<!--\; -\; -->1)\left(\frac{3}{4}\right)$

$=-<!--\; -\; -->129-<!--\; -\; -->\frac{1}{4}+15-<!--\; -\; -->\frac{3}{4}$

$=-<!--\; -\; -->115$