Find the geometric progression with fourth term = 54 and seventh term = 1458.

Let first term of the G.P. be a and it's common ratio be r.

Given,

⇒ a₄ = 54

⇒ ar³ = 54 ……..(i)

Also,

⇒ a₇ = 1458

⇒ ar⁶ = 1458 ……..(ii)

Dividing (ii) by (i) we get

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

Substituting value of r in (i) we get,

⇒ a(3)³ = 54

⇒ 27a = 54

⇒ a = $\dfrac{54}{27}$ = 2.

a₂ = ar

= 2.(3) = 6.

a₃ = ar²

= 2.(3)²

= 2.(9) = 18.

G.P. = 2, 6, 18, 54, ……

Hence, G.P. = 2, 6, 18, 54, ……