What is the 8th term of the geometric sequence where a₁ = 256 and a₃ = 16?

Solution:

The nth term of a geometric sequence is given by

a_n = a₁ × r^{n - 1} --- (1)

Given,

a₁ = 256, a₃ = 16

To find r put the value of a₁ and a₃ in (1)

⇒ 16 = 256 × r²

⇒ r² = 16/256

⇒ r² = 0.0625

⇒ r = 0.25

Now, find a₈

a₈ = a₁ × r⁷

a₈ = 256 × (0.25)⁷

= 256 × 0.00006104

a₈ = 0.0156

Therefore, the 8th term of the geometric sequence is 0.0156

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What is the 8th term of the geometric sequence where a₁ = 256 and a₃ = 16?

Summary:

The 6th term of the geometric sequence where a₁ = 256 and a₃ = 16 is 0.0156