$α, β, γ$ are the geometric means between $c a, a b; a b, b c; b c, c a$ respectively. Prove that if $a, b, c$ are in A.P., then $α^{2}, β^{2}, γ^{2}$ are also in A.P., and $β + γ, γ + α, α + β$ are in H.P.

Open in App

Solution

By given condition
$α^{2} = a^{2} b c, β^{2} = b^{2} c a, γ^{2} = c^{2} a b a n d 2 b = a + c$
$α^{2}, β^{2}, γ^{2}$ will be in A.P. if $a^{2} b c, b^{2} c a, c^{2} a b$
are in A.P. or a,b,c are in A.P. which is true.
Again $β + γ, γ + α, α + β$ will be in H.P. if
$\frac{1}{β + γ}, \frac{1}{γ + α}, \frac{1}{α + β}$
or $\frac{1}{γ + α} - \frac{1}{β + γ} = \frac{1}{α + β} - \frac{1}{γ + α}$
or $\frac{β - α}{β - γ} = \frac{γ - β}{α + β} o r β^{2} - α^{2} = γ^{2} - β^{2}$
or $α^{2}, β^{2}, γ^{2}$ are in A.P. Which we have already proved.

Suggest Corrections

Similar questions

α, β, γ

α^{2}, β^{2}, γ^{2}

α, β, λ

a, b, c

A . P

α^{2}, β^{2}, λ^{2}

α

β

γ

x^{3} + q x + r = 0

\frac{β + γ}{α^{2}}, \frac{γ + α}{β^{2}}, \frac{α + β}{γ^{2}}

α, β, γ

x^{3} + x^{2} + x + 1 = 0

(α - β)^{2} + (β - γ)^{2} + (γ - α)^{2}

α, β, γ

α + β + γ = 2, α^{2} + β^{2} + γ^{2} = 6, α^{3} + β^{3} + γ^{3} = 8

α^{4} + β^{4} + γ^{4}

Join BYJU'S Learning Program