The sequence is geometric, so
[tex]a_n = r a_{n-1}[/tex]
for some constant r. From this rule, it follows that
[tex]a_3 = r a_2 \implies 20 = 2r \implies r = 10[/tex]
and we can determine the first term to be
[tex]a_2 = r a_1 \implies 2 = 10 a_1 \implies a_1 = \dfrac15[/tex]
Now, by substitution we have
[tex]a_n = r a_{n-1} = r^2 a_{n-2} = r^3 a_{n-3} = \cdots[/tex]
and so on down to (D)
[tex]a_n = r^{n-1} a_1 = 10^{n-1} \cdot \dfrac15[/tex]
(notice how the exponent on r and the subscript on a add up to n)
