Answer: [tex]\bold{a_n=\dfrac{2^{n-1}}{3^{n+1}}}[/tex]
Step-by-step explanation:
The explicit rule for a geometric sequence is: [tex]a_n=a_1(r)^{n-1}\quad \text{where}\ a_1\ \text{is the first term and r is the common ratio}[/tex]
Given the sequence {9, 6, 4, [tex]\dfrac{8}{3}[/tex], ... } we know
- the first term (a₁) is 9
- the common ratio (r) is [tex]\dfrac{6}{9}=\dfrac{2}{3}\ \text{when simplified}[/tex]
So, the explicit rule for the given sequence is:
[tex]a_n=9\bigg(\dfrac{2}{3}\bigg)^{n-1}\\\\\\.\quad =3\cdot3\dfrac{(2)^{n-1}}{(3)^{n-1}}\\\\\\.\quad =3\dfrac{(2)^{n-1}}{(3)^{n}}\\\\\\.\quad =\dfrac{(2)^{n-1}}{(3)^{n+1}}[/tex]
