Find the requested term of the geometric sequence?

#1 ; a = 2 

#5 ; =ar^4 = 512 

Divide

r^4 = 256 

r = 256^(1/4) 

r = 4 

Hence #3 ; ar^2 = 2(4)^2 = 32 

For a geometric progression:

a₁

a₂ = q * a₁ → where q is the common ratio

a₃ = q * a₂ = q² * a₁

a₄ = q * a₃ = q³ * a₁

a₅ = q * a₄ = q⁴ * a₁

The third term of the geometric sequence is 2: → a₃ = q² * a₁ = 2

The fifth term of a geometric sequence is 512: → a₅ = q⁴ * a₁ = 512

a₅/a₃ = (q⁴ * a₁)/(q² * a₁) = 512/2

(q⁴ * a₁)/(q² * a₁) = 512/2

q⁴/q² = 256

q² = 256

q = ± 16

Recall: a₃ = 2

q² * a₁ = 2

a₁ = 2/q²

a₁ = 2/256

a₁ = 1/128

a₂ = q * a₁

a₂ = ± 16 * (1/128)

a₂ = ± 1/8

a₃ = q² * a₁

a₃ = 256 * (1/128)

a₃ = 2

a₄ = q³ * a₁

a₄ = (± 16)³ * (1/128)

a₄ = ± 4096/128

a₄ = ± 32

an = a₁r^(n-1)

a₁ = 2

a5 = 2r^(5-1) = 2r^4 = 512

r = 256^(1/4) = 4

an = 2(4)^(n-1)

a3 = 2(4)^(3-1) = 2(4²) = 2(16) = 32

 Find the third term of a geometric sequence 

  whose first term is 2 and whose fifth term is 512.

  2r^4 = 512

  r^4 = 256

  r = 4

  2, 8, 32, 128, 256, ...

  The third term is 32.

Geometric = next term = this term multiplied by a constant number.

In this case, it looks like

n=1, term = 2

n=5, term = 2*4*4*4*4 = 2^9

I suspect that

n=2, term = 2*4 = 2^3

The constant factor is 4