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$x_1,x_2,x_3,x_4$ are in Harmonic Progression I need to show $$(x_1-x_3)(x_2-x_4)=4(x_1-x_2)(x_3-x_4)$$

I tried assuming reciprocals are in Arithmetic Progression, but after huge calculation I did not come any result.

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    $\begingroup$ reciprocal solves the problem, just divide on both sides $x_1x_2x_3x_4$. $\endgroup$
    – Troy Woo
    Aug 13 '14 at 17:59
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Given that $x_1$, $x_2$, $x_3$ and $x_4$ are terms of a harmonic progression, we can denote $x_1=\frac{1}{a}$, $x_2=\frac{1}{a+d}$, $x_3=\frac{1}{a+2d}$, $x_4=\frac{1}{a+3d}$, for constants $a$ and $d$, the arithmetic progression difference.

Expressing $x_k$ in terms of $a$ and $d$, we have $$(x_1-x_3)(x_2-x_4)=\frac{2d}{a(a+2d)}\frac{2d}{(a+d)(a+3d)}\\=4\color{red}{\frac{d}{a(a+d)}}\color{blue}{\frac{d}{(a+2d)(a+3d)}}\\=4\color{red}{(x_1-x_2)}\color{blue}{(x_3-x_4)}$$

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Since $x_1,x_2,x_3,x_4$ are in harmonic progression then:

$$x_1 = \frac{1}{a} \hspace{1cm}x_2 = \frac{1}{a+d} \hspace{1cm}x_3 = \frac{1}{a+2d} \hspace{1cm}x_4 = \frac{1}{a+3d}$$

Therefore LHS:

$$(x_1-x_3)(x_2-x_4) = \left(\frac{1}{a}-\frac{1}{a+2d}\right)\left(\frac{1}{a+d}-\frac{1}{a+3d}\right) = \\ = \frac{a+2d-a}{a(a+2d)}·\frac{a+3d-a-d}{(a+d)(a+3d)} = \frac{4d^2}{a(a+d)(a+2d)(a+3d)} = 4d^2x_1x_2x_3x_4$$

And RHS:

$$4(x_1-x_2)(x_2-x_3) = 4\left(\frac{1}{a}-\frac{1}{a+d}\right)\left(\frac{1}{a+2d}-\frac{1}{a+3d}\right) = \\ = 4\frac{a+d-a}{a(a+d)}·\frac{a+3d-a-2d}{(a+2d)(a+3d)} = 4\frac{d^2}{a(a+d)(a+2d)(a+3d)} = 4d^2x_1x_2x_3x_4$$

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