Question

if a1, a2, a3 (a1>0) are three successive terms of a G.P with common ratio r, the value of r for which the value of r for which a3 > 4a2 - 3a1 holds is given by​

Answer 1

Given info : a₁, a₂ and a₃ (a₁ > 0) are three successive terms of a GP with common ratio r.

To find : for the value of r for which a₃ > 4a₂ - 3a₁

Solution : as a₁ , a₂ and a₃ are in geometric progression and common ratio is r.

so, a₂ = a₁r , a₃ = a₁r²

Now, a₃ > 4a₂ - 3a₁

⇒a₁r² > 4a₁r - 3a₁

⇒a₁(r² - 4r + 3) > 0

⇒(r² - 4r + 3) > 0

⇒(r² - 3r - r + 3) > 0

⇒(r - 3)(r - 1) > 0

⇒r > 3 or r < 1

Therefore the value of r ∈ (3, ∞) U (-∞ , 1)