3, 8, 13, ……253 are in A.P, then

Given 3, 8, 13, ….253 are in A.P

Here $a=3,d=,l=t_n=253$

$a+(n-1)d=253$

$3+(n-1)5=253$

$n-1=\frac{250}{5}$

$n=51$

20^th term from end is ${(n-p+1)}^{th}$ term from beginning

$\therefore t_{(n-p+1)}=t_{(51-20+1)}=t_{32}=a+31d$

$=3+31\left(5\right)=3+155=158$

18^th term from end is ${(n-p+1)}^{th}$ term from beginning

$\therefore t_{(n-p+1)}=t_{(51-18+1)}=t_{34}=a+33d$

$=3+33\left(5\right)=3+165=168$