Find the geometric progression whose 4th term is 54 and 7th term is 1458.

Given, a₄ = 54 and a₇ = 1458.

We know that in G.P.

   a_n = ar^{n - 1}
∴ a₄ = ar³ and a₇ = ar⁶.

Dividing a₇ by a₄,

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \therefore r = 3. \\[1em]$

Putting value of r in ar³ = 54,

$\Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2. \\[1em]$

Terms of G.P. are

⇒ a₂ = ar = 2 × 3 = 6,
⇒ a₃ = ar² = 2 × 3² = 18,
⇒ a₄ = ar³ = 2 × 3³ = 54.

Hence, the G.P. is 2, 6, 18, 54, ….