Question

If a is the A.M. of b and c and the two geometric means are G₁ and G₂, then \(G_1^3\) + \(G_2^3\) is equal to 

(a) \(\frac{abc}{2}\)

(b) abc 

(c) 2abc 

(d) \(\frac{3}{2}abc\)

Answer 1

(c) 2abc

a is the A.M of b and c ⇒ 2a = b + c 

Given, G₁ and G₂ are G.Ms between b and c 

⇒ b, G₁, G₂, c are in G.P. 

Let r be the common ratio of the G.P. 

⇒ G₁ = br, G₂ = br², c = br³ 

Now c = br³ ⇒ r³ = \(\frac{c}{b}\) ⇒ r = \(\big(\frac{c}{b}\big)^\frac{1}{3}\)

∴ G₁ = b x \(\frac{c^\frac{1}{3}}{b^\frac{1}{3}}\) = \(b^\frac{2}{3}\) \(c^\frac{1}{3}\) = \((b^2c)^\frac{1}{3}\)

G₂ = b x \(\big(\frac{c}{b}\big)^\frac{2}{3}\) = \(\frac{b\times{c}^\frac{2}{3}}{b^\frac{2}{3}}\) = \(b^\frac{1}{3}\) \(c^\frac{2}{3}\) = \((bc^2)^\frac{1}{3}\)

Now \(G_1^3\) + \(G_2^3\) = \(\big((b^2c)^\frac{1}{3}\big)^3\) + \(\big((bc^2)^\frac{1}{3}\big)^3\)

= b²c + bc² = bc (b + c) = bc . 2a = 2abc.