Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.

Given: a₃ = a₁ + 9 and a₂ = a₄ + 18

Let a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³

Here,

a₃ = a₁ + 9

⇒ ar² = a + 9

⇒ ar² – a = 9

⇒ a(r² – 1) = 9 – 1

and,

a₂ = a₄ + 18

⇒ ar= ar³ + 18

⇒ ar – ar³ = 18

⇒ ar(1 – r²) = 18 – 2

Divide eq –2 by eq –1

⇒

⇒

⇒ r = – 2

Substitute r in eq—1

a × ((-2)² – 1) = 9

a × (4 – 1) = 9

a × (3) = 9

a = = 3

∴ G.P is : 3 , 3(– 2), 3(– 2)², 3(– 2)³

⇒ 3, —6 , 12 , —16

∴ G.P is 3, —6, 12, —16