Insert 6 geometric means between 27 and 1/81.

Let the six terms be a₁, a₂, a₃, a₄, a₅, a₆.

A = 27, B = \( \dfrac{1}{81}\)

Now, these 6 terms are between A and B.

So the GP is: A, a₁, a₂, a₃, a₄, a₅, a₆, B.

So we now have 8 terms in GP with the first term being 27 and eighth being \( \dfrac{1}{81}\).

We know that, T_n = ar^n–1

Here, T_n = \( \dfrac{1}{81}\), a = 27 and

\( \dfrac{1}{81}\)= 27r^8-1

\(\dfrac{ 1}{(81×27)}\) = r⁷

r = \( \dfrac{1}{3}\)

a₁ = Ar = 9

a₂ = Ar² = 3

a₃ = Ar³ = 1

a₄ = Ar⁴ = \( \dfrac{1}{3}\)

a₅ = Ar⁵ = \( \dfrac{1}{9}\)

a₆ = Ar⁶ = \( \dfrac{1}{27}\)

The six GM between 27 and \( \dfrac{1}{81}\) are 9, 3, 1, \( \dfrac{1}{3}\),\( \dfrac{1}{9}\), \( \dfrac{1}{27}\)