Question

The 4th term of a geometric sequence is -512 and the 9th term is 16 find the 15th term 

Answer 1

a₁₅ =
`(1)/(4)`

Explanation:

the nth term of a geometric sequence is

`a_(n)` = a₁
`(r)^(n-1)`

where a₁ is the first term and r the common ratio

we require to fid a₁ and r

given a₄ = - 512 and a₉ = 16 , then

a₁ r³ = - 512 → (1)

a₁
`r^(8)` = 16 → (2)

divide (2) by (1)

`(a_(1)r^(8) )/(a_(1)r^(3) )` =
`(16)/(-512)` ( cancel a₁ on numerator/ denominator on left side )

`(r^(8) )/(r^(3) )` = -
`(1)/(32)`

`r^(5)` = -
`(1)/(32)`

r =
`\sqrt[5]{-(1)/(32) }` = -
`(1)/(2)`

substitute r = -
`(1)/(2)` into (1) and solve for a₁

a₁ × (-
`(1)/(2)` )³ = - 512

a₁ × -
`(1)/(8)` = - 512 ( multiply both sides by - 8 to clear the fraction )

a₁ = 4096

Then

a₁₅ = 4096 ×
`(-(1)/(2)) ^(14)` = 4096 ×
`(1)/(16384)` =
`(4096)/(16384)` =
`(1)/(4)`