If you have that $a_n=a_{n-1}+5$, with $a_1=3$, then you have that $a_n=5(n-1)+3$, because $$a_n=a_{n-1}+5=(a_{n-2}+5)+5=a_{n-2}+10=(a_{n-3}+5)+10=a_{n-3}+15=$$$$...=a_1+5(n-1)$$
If you have that $a_n=a_{n-1}+5$, with $a_1=3$, then you have that $a_n=5(n-1)+3$, because $$a_n=a_{n-1}+5=(a_{n-2}+5)+5=a_{n-2}+10=(a_{n-3}+5)+10=a_{n-3}+15=$$$$...=a_1+5(n-1)$$