According to the given problem,
[tex]\begin{gathered} a_n=a_{n-1}+9 \\ a_1=10 \end{gathered}[/tex]Solve for the successive terms by substituting n=2,3,4,5,....
[tex]\begin{gathered} a_2=a_1+9=10+9=19 \\ a_3=a_2+9=19+9=28 \\ a_4=a_3+9=28+9=37 \\ a_5=a_4+9=37+9=46 \end{gathered}[/tex]So the sequence is obtained as,
[tex]10,19,28,37,46,\ldots\ldots\ldots[/tex]Onserve that this is an Arithmetic Progression with first term (a) 10, and the common difference (d) 9 units.
So the nth term of the AP is given by,
[tex]\begin{gathered} a_n=a+(n-1)d \\ a_n=10+(n-1)9 \\ a_n=10+9n-9 \\ a_n=1+9n \end{gathered}[/tex]This the explicit formula representing the sequence is obtained as,
[tex]a_n=1+9n[/tex]