[tex]A(n)=5+(n-1)\left(\dfrac{1}{6}\right)=5+\dfrac{1}{6}n-\dfrac{1}{6}=\dfrac{1}{6}n+\dfrac{30}{6}-\dfrac{1}{6}\\\\A(n)=\dfrac{n+29}{3}\\\\\text{Put n = 1, n = 4 and n = 10 to the formula:}\\\\A(1)=\dfrac{1+29}{3}=\dfrac{30}{3}=10\\\\A(4)=\dfrac{4+29}{3}=\dfrac{33}{3}=11\\\\A(10)=\dfrac{10+29}{3}=\dfrac{39}{3}=13[/tex]
