Answer:
C
Step-by-step explanation:
There is a common ratio between consecutive term in the sequence, that is
[tex]\frac{3}{10}[/tex] ÷ [tex]\frac{3}{100}[/tex] = 3 ÷ [tex]\frac{3}{10}[/tex] = 30 ÷ 3 = 300 ÷ 30 = 10
This indicates the sequence is geometric with n th term
[tex]a_{n}[/tex] = a[tex](r)^{n-1}[/tex]
where a is the first term and r the common ratio
Here a = [tex]\frac{3}{100}[/tex] and r = 10 , thus
[tex]a_{n}[/tex] = [tex]\frac{3}{100}[/tex] [tex](10)^{n-1}[/tex]
= [tex]\frac{3}{10^{2} }[/tex] × [tex]10^{n-1}[/tex]
= 3 × [tex]10^{-2}[/tex] × [tex]10^{n-1}[/tex]
= 3 × [tex]10^{n-3}[/tex]
Thus
[tex]a_{n}[/tex] = 3[tex](10)^{n-3}[/tex] → C
