Determine the AP whose fifth term is 19

Let the first term of an AP be a and common difference d.

Given, a5 = 19and a13 – a8 = 20         [given]

$\therefore a_{5}=a+(5-1) d=19$ and $[a+(13-1) d]-[a+(8-1) d]=20 \quad\left[\because a_{n}=a+(n-1) d\right]$

$\Rightarrow \quad a+4 d=19 \quad \ldots$ (i)

and $\quad a+12 d-a-7 d=20 \Rightarrow 5 d=20$

$\therefore$ $d=4$

On putting $d=4$ in Eq. (i), we get

$a+4(4)=19$

$a+16=19$

$a=19-16=3$

So, required AP is $a, a+d, a+2 d, a+3 d, \ldots$ i.e., $3,3+4,3+2(4), 3+3(4), \ldots$ i.e., $3,7,11,15, \ldots$