What is the 7th term of the geometric sequence where a₁ = 128 and a₃ = 8?

Solution:

The nth term of the geometric sequence is given by, a_n = a r^{n - 1},

Where a and r being the first term and the common ratio respectively.

Given a₁ = a = 128 and a₃ = 8

We know a_n = a r^{n - 1},

⇒ a₃ = a.r^{3 - 1}

⇒ 8 = 128 × r²

⇒ r² = 8/128

⇒ r² = 1/16

⇒ r = ±1/4

Now, a₇ = a.r^{7 - 1} = a.r⁶

⇒ a₇ = 128 × (1/4)⁶

⇒ a₇ = 128/4096

⇒ a₇ = 1/32

If r = -1/4,

⇒ a₇ = 128 × (-1/4)⁶

⇒ a₇ = 128/4096

⇒ a₇ = 1/32

Therefore, the 7th term of the geometric sequence a₇ is 1/32.

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What is the 7th term of the geometric sequence where a₁ = 128 and a₃ = 8?

Summary:

The 7th term of the geometric sequence where a₁ = 128 and a₃ = 8 is 1/32.