Answer:
4, 8, 16, 32, 64
Step-by-step explanation:
The nth term of a geometric sequence is
[tex]a_{n}[/tex] = a₁[tex](r)^{n-1}[/tex]
Given
a₇ = 256 and a₁₀ = 2048 , then
a₁ [tex]r^{6}[/tex] = 256 → (1)
a₁ [tex]r^{9}[/tex] = 2048 → (2)
Divide (2) by (1)
[tex]\frac{a_{1}r^{9} }{a_{1}r^{6} }[/tex] = [tex]\frac{2048}{256}[/tex]
r³ = 8 ( take the cube root of both sides )
r = [tex]\sqrt[3]{8}[/tex] = 2
Substitute r = 2 into (1)
a₁ × [tex]2^{6}[/tex] = 256
a₁ × 64 = 256 ( divide both sides by 64 )
a₁ = 4
Then
a₁ = 4
a₂ = 2a₁ = 2 × 4 = 8
a₃ = 2a₂ = 2 × 8 = 16
a₄ = 2a₃ = 2 × 16 = 32
a₅ = 2a₄ = 2 × 32 = 64
