If a is A.M. of b and c and the two geometric means are G1 and G2, the

1.	If a is A.M. of b and c and the two geometric means are G1 and G2, then prove that \(G^3_1\) + \(G^2_3\) = 2abc
Answer» It is given that a is the A.M. ot b and c. ∴ a = \(\frac{b + c}{2}\) ⇒ b + c = 2a … (i) Since G₁ and G₂ are two geometric means between b and c. Therefore, b, G₁, G₂ c is a G.P. with common ratio r = \(\big(\frac{c}{b}\big)^\frac{1}{3}\) ∴ G₁ = br = b\(\big(\frac{c}{b}\big)^\frac{1}{3}\) = \(C^\frac{1}{3}b^\frac{1}{3}\) and G₂ = br² = b\(\big(\frac{c}{b}\big)^\frac{2}{3}\) = \(b^\frac{1}{3}b^\frac{2}{3}\) ⇒ \(G_1^3\) = b²c and ⇒ \(G_2^3\) = bc² ⇒ \(G_1^3\) + \(G_2^3\) = b²c bc² ⇒ \(G_1^3\) + \(G_2^3\) = bc(b + c) ⇒ \(G_1^3\) + \(G_2^3\) = 2abc [using (i)] Hence proved

If a is A.M. of b and c and the two geometric means are G1 and G2, then prove that \(G^3_1\) + \(G^2_3\) = 2abc

Answer»

It is given that a is the A.M. ot b and c.

∴ a = \(\frac{b + c}{2}\) ⇒ b + c = 2a … (i)

Since G₁ and G₂ are two geometric means between b and c. Therefore, b, G₁, G₂ c is a G.P.

with common ratio r = \(\big(\frac{c}{b}\big)^\frac{1}{3}\)

∴ G₁ = br = b\(\big(\frac{c}{b}\big)^\frac{1}{3}\) = \(C^\frac{1}{3}b^\frac{1}{3}\) and

G₂ = br² = b\(\big(\frac{c}{b}\big)^\frac{2}{3}\) = \(b^\frac{1}{3}b^\frac{2}{3}\)

⇒ \(G_1^3\) = b²c and ⇒ \(G_2^3\) = bc²

⇒ \(G_1^3\) + \(G_2^3\) = b²c bc² ⇒ \(G_1^3\) + \(G_2^3\) = bc(b + c)

⇒ \(G_1^3\) + \(G_2^3\) = 2abc

[using (i)]

Hence proved