What is the 6th term of the geometric sequence where a₁ = 128 and a₃ = 8?

Solution:

The nth term of a geometric sequence is given by

a_n = a₁ × r^{n - 1} --- (1)

Given, a₁ = 128, a₃ = 8

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

⇒ 8 = 128 × r²

⇒ r² = 8/128

⇒ r² = 0.0625

⇒ r = 0.25

Now, find a₆

a₆ = a₁ × r⁵

a₆ = 128 × (0.25)⁵

= 128 × 0.000976

a₆ = 0.125

Therefore, the 6th term of the geometric sequence is 0.125

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What is the 6th term of the geometric sequence where a₁ = 128 and a₃ = 8?

Summary:

The 6th term of the geometric sequence where a₁ = 128 and a₃ = 8 is 0.125