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

Given, a₄ = 54 and a₇ = 1458.

By formula, a_n = ar^{n - 1}.

⇒ a₄ = a(r)^{4 - 1}
⇒ 54 = ar³     (Eq 1)

⇒ a₇ = a(r)^{7 - 1}
⇒ 1458 = a(r)⁶     (Eq 2)

Dividing Eq 2 by Eq 1,

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

Putting value of r in Eq 1,

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

a₂ = ar = 2 × 3 = 6
a₃ = ar² = 2(3)² = 2 × 9 = 18
a₄ = ar³ = 2(3)³ = 2 × 27 = 54.

Hence, the required G.P. is 2, 6, 18, 54, …