If m is the A.M. of two distinct real numbers ℓ and n (ℓ, n > 1) and G_1, G_2 and G_3 are three geometric means between ℓ and n, then G_1^4 + 2G_2^4 + G_3^4 equals:

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EEJ MAIN Mathematics QUESTION #7491

Question 1

If m is the A.M. of two distinct real numbers ℓ and n (ℓ, n > 1) and \(G_1, G_2\) and \(G_3\) are three geometric means between ℓ and n, then \(G_1^4 + 2G_2^4 + G_3^4\) equals:

\(4\ell^2 mn\)
\(4\ell m^2 n\)✔️
\(4\ell mn^2\)
\(4\ell^2 m^2 n^2\)

Correct Answer Explanation

We have \(m = \frac{\ell+n}{2}\). For three GMs between \(\ell\) and \(n\): \(\ell, G_1, G_2, G_3, n\) with common ratio \(r = (\frac{n}{\ell})^{1/4}\). Thus \(G_2 = \ell r^2 = \ell\sqrt{\frac{n}{\ell}} = \sqrt{\ell n}\), so \(G_2^2 = \ell n\). Also, \(G_1 G_3 = (\ell r)(\ell r^3) = \ell^2 r^4 = \ell n\). We have \(G_1^2 + G_3^2 = (\ell r)^2 + (\ell r^3)^2 = \ell^2 r^2(1+r^4) = \ell^2(\frac{n}{\ell})^{1/2}(1+\frac{n}{\ell})\). Through algebraic manipulation, the expression evaluates to \(4\ell m^2 n\).

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