$\mathrm{If} \mathrm{the} 10\mathrm{th} \mathrm{term} \mathrm{of} \mathrm{an} \mathrm{AP} \mathrm{is} 52 \mathrm{and} 17\mathrm{th} \mathrm{term} \mathrm{is} 20 \mathrm{more} \mathrm{than} \mathrm{the} 13\mathrm{th} \mathrm{term}, \mathrm{find} \mathrm{the} \mathrm{AP}.$

$We have a_{10}=52 and a_{17}=a_{13}+20$$\Rightarrow a+9d=52 and a+16d=a+12d+20$

$\Rightarrow a+9d=52 and 4d=20$

$\Rightarrow a+9d=52 and d=5$

$\Rightarrow a+9\left(5\right)=52$

$\Rightarrow a+45=52$

$$\begin{gathered} \Rightarrow a=7 \\ Hence the AP is 7,12,17,22,\dots \end{gathered}$$