InterviewSolution
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InterviewSolution
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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) \(\frac{3}{2}abc\) |
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Answer» (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. |
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