Answer:
[tex]a_{n}[/tex] = 2[tex](-4)^{n-1}[/tex]
Step-by-step explanation:
The n th term of a geometric sequence is
[tex]a_{n}[/tex] = a[tex](r)^{n-1}[/tex]
where a is the first term and r the common ratio
Given
[tex]a_{2}[/tex] = - 8, then ar = - 8 → (1)
Given
[tex]a_{5}[/tex] = 512, then a[tex]r^{4}[/tex] = 512 → (2)
Divide (2) by (1)
[tex]\frac{ar^4}{ar}[/tex] = [tex]\frac{512}{-8}[/tex], thus
r³ =- 64 ( take the cube root of both sides )
r = [tex]\sqrt[3]{-64}[/tex] = - 4
Substitute r = - 4 into (1)
a(- 4) = - 8
- 4a = - 8 ( divide both sides by - 4 )
a = 2
Thus
[tex]a_{n}[/tex] = 2[tex](-4)^{n-1}[/tex]
