Answer:
Option A is correct.
[tex]a_n =-15 (\frac{1}{5})^{n-1}[/tex]
Step-by-step explanation:
Given: [tex]a_1 = -15[/tex] ; [tex]a_n = \frac{1}{5} \cdot a_{n-1}[/tex]
for n = 2
[tex]a_2 = \frac{1}{5} \cdot a_{2-1}[/tex] = [tex]\frac{1}{5}\cdot a_1 = \frac{1}{5} \cdot (-15)[/tex]
for n =3
[tex]a_3 = \frac{1}{5} \cdot a_{3-1}[/tex] = [tex]\frac{1}{5} \cdota_2= (\frac{1}{5})^2\cdot (-15)[/tex]
Simlarly , for n =4
[tex]a_3 = \frac{1}{5} \cdot a_{4-1}[/tex] = [tex]\frac{1}{5}\cdot a_3= (\frac{1}{5})^3\cdot (-15)[/tex]
and so on...
Common ratio(r) states that for a geometric sequence or geometric series, the common ratio is the ratio of a term to the previous term.
we have; [tex]r = \frac{a_2}{a_1} = \frac{1}{5}[/tex]
Now; by recursive formula for geometric series:
[tex]a_n = a_1 r^{n-1}[/tex]
where [tex]a_1[/tex] is the first term and r is the common ratio:
Substitute the value of [tex]a_1 = -15[/tex] and [tex]r = \frac{1}{5}[/tex]
we have;
[tex]a_n =-15 (\frac{1}{5})^{n-1}[/tex]
therefore, the explicit rule for the geometric sequence is; [tex]a_n =-15 (\frac{1}{5})^{n-1}[/tex]
