Let the first term of an AP be "a" and common difference be "d".
Given, a5=19
⇒a5=a+(5−1)d=19 [∵an=a+(n−1)d]
and a13−a8=20
⇒[a+(13−1)d]−[a+(8−1)d]=20 [∵an=a+(n−1)d]
⇒a+4d=19...(i)
⇒a+12d−a−7d=20
⇒5d=20
∴d=4
Substituting d = 4 in eq. (i), we get;
a + 4(4) = 19
a + 16 = 19
a = 19 - 16 = 3
So, the required AP is : a, a+d, a+2d, a+3d, ...
= 3, 3+4, 3+2(4), 3 +3 (4), …
= 3, 7, 11, 15, ...